i love watching this guy.. i used to have to work things like this out on paper years ago as a mechanical precision , but then i worked for other companies where they have auto cad. so i didn't need to work it out anymore, because the guys gave me the answers,.. well done teacher.. i love watching your videos.
So nice of you Paul! You are awesome 👍 Thank you so much for taking the time to leave this comment. I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
I am also lazy in my old age, I can do this the way shown but I just timed myself using CAD - drawing the three lines, drawing a circle from three points and finding the diameter of the circle. It took 90 seconds and that included starting the CAD program up. It's just another tool like a pen and paper, slide rule or calculator. We all take the easy way out if we can.
@@xoxb2 yeah atleast he isn't bragging about how much time it took reducing all the steps he took to make idiots like that guy understand he would probably solve in half his time
Put A to coordinate (0,0), B to (0,4) and C to (6,6). Put the coordinates into the equation of the circle. You get 3 unknown (coordinates for circle center and r) and 3 equations. Solve for (x,y) för center of circle and radius. If you have the coordinates for 3 points on the circle you can draw the circle. More algebra, less geometry.
Even easier: Let the circle center be (0,0), then A =(x,-2) and B=(x,2) because of symmetry. C=(x+6,4). |C|^2=r^2=|B|^2, which reduces to a linear equation in x.
Without any geometry, you can find the radius using coordinates and a little algebra. Put the origin at the point A (0,0), then B is (0,4) and C is (6,6). You know the equation of the circle is of the form (x+a)^2 + (y+b)^2 = r^2 Substituting x and y from the three points gives three equations: a^2 + b^2 = r^2 a^2 + b^2 + 8b + 16 = r^2 a^2 + b^2 + 12a + 12b + 36 +36 = r^2 You can subtract the first equation from each of the second and third equations giving: 8b + 16 = 0 12a +12b +72 = 0 The former gives: b = -2 and then using that in the latter gives: 12a - 24 + 72 = 0 which leads to a = -4 Since r^2 = a^2 + b^2, we get: r^2 = 16 + 4 = 20 so r = sqrt(20) or 2.sqrt(5) ~= 4.472 No constructions; no "by symmetry"; no intersecting chords; no angle in a semicircle; no Pythagoras, just a pair of linear simultaneous equations.
My solution as well. Cut to the crap. But as an old math teacher it is nice to see the connection btw pythagoras, Thales, corda. I would show both methods. If I still worked as a teacher.
I did something similar, except I didn't move the origin, I only rotated the circle until the cord ab was perfectly vertical. so my points were a(x,-2) b(x,2) and c(x+6, 4) substituting a and c into the equation we get x^2 + 4 =x^2 + 12x +52 solve for x => 4 = 12x + 52 => 12x = 4-52 => x = -48/12 = -4 we plug our variables back into the circle equation and solve for r, we get r = sqrt(20) = 2sqrt(5)
Another solution. Let OA=OB=OC=r. Draw a perpendicular line from point O to line AB, and let the intersection point be E. Let the length of line EO be x. Since triangle OAB is an isosceles triangle (a) r^2 = 2^2 + x^2 Extend line CD and line EO, and set the intersection point to F. Since the length of the line segment CF will be 4 (b) r^2 = 4^2 + (6-x)^2 From (a) and (b) 2^2 + x^2 = 4^2 + (6-x)^2 Therefore x=4 Substituting x=4 into (a), we get r=2 Sqrt (5)
If you went to school and never heard of them what makes you think going back would help? If you heard about them on your own why do not seek self study instead?
Another commenter pointed out that because of symmetry due to the inscribed rectangle What inscribed rectangle?, I asked myself. What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
Thanks, I finally got it when I realized that one edge of the square or rectangle had to be a chord. Seems obvious now. Love math, subscribed! And please pardon the double post. :)
Once you know all 4 chord segments, you can use 4r²=CD² + DF² + BD² + DE². The reason why the Pythagorean theorem works in this case is an inscribed angle in a circle is 1/2 its intercepted arc. Since angle ABE is 90 degrees arc AFE is 180 degrees. When the line is drawn from A to E, it must pass through the center of the circle.
I solved it this way. *Draw line segment AC and using similar triangles we can prove angle BAC is 45° *Next mark center O and draw triangle BOC. Angle BOC is 90° using arc-centre degree measure theorem *BC can be computed by applying Pythagorean theorem on BDC *Now in right triangle BOC, apply Pythagorean theorem to find out the radius of the circle
I think you applied the converse of Thales' Theorem: If A, B, and C are points on a circle where the angle ACB is a right angle, then the line AB is a diameter of the circle. Of course, this has also been proven true.
Exactly. That's what I came here to say. The fact that A --> B does NOT by itself mean that B --> A. In fact, simply claiming that it does is a classic logical fallacy (affirming the consequent.) So this was not a proper application of Thales' Theorem. Now, it appears that the converse of Thales' Theorem is in fact ALSO true, but it should be treated as a separate theorem, to make it clear that the solution of the problem here is not proceeding illogically. It needs to be stated categorically that B --> A independently.
Nice, I've got the right answer without knowing any of those theorems. I've just calculated the radius using phythagoras with r = √(2^2+4^2). With 2 beeing half the lenghth of the left line (symmetry) and 4 beeing the lenghth of the horizontal from the point on the circle to the intersection of the horizontal line and the conecting line from the point on the circle on the bottom and the the point on the circle on the top.
If you draw a parallel to BE that goes through A, then CF is divided into three parts that must - according to symmetry - have the length 2 - 4 - 2. DF therefore has a length of 6 and is therefore the same length as BD. Therefore CD is equal to DE. So we have a rectangle inside the circle with the corners A, B, E and the second intersection of the drawn parallel with the circle. The rectangle has side lengths 8 and 4. Thus, the diagonal of the rectangle is d, and we have: d² = 4² + 8² = 16 + 64 = 80 r = 1/2 * d and thus r² = 80 / 4 = 20 r is therefore 2 * sqrt(5) = 4.472 long.
My method was quite different. You have the cord AB and can make a second cord BC. We know that a perpendicular line from the mid point of a cord passes through the centre of the circle. The line BC has a slope 2/6 =1/3 relative to the horizontal. The centre of BC will be 3 to the right from B and 1 above B. The centre of the circle will be 3 below this point. The slope of the perpendicular line from BC 1/3 relative to the vertical. The centre of the circle will therefore be 1 to right of the centre of BC. The centre will be 3+1 =4 to the right of AB. Half of AB = 2. Pythagoras time. R^2 = 4^2 + 2^2 = 16 + 4 =20 R = root 20 =2 root 5
Indeed. The center of the circle will be on a line parallel to and 2 below of AD; that line is perpendicular to AB and cuts it in half. It will also be on the perpendicular line that cuts AC in half; it cuts that line 1 above AD. As AC is sloped 3/1 it is sloped 1/3, and as the center is 3 lower it is one to the right of the middle of AC, so 3+1 = 4 to the right of AB. Then r = √(4² + 2²) = √20 = 2√5.
That's much cleaner than how I often do it. Once I have calculated chord lengths, I might move a chord to become the diameter. For instance, moving BE down to cross the centre would have it as (x+6)(x+2) where x is the distance between chord end and circumference (it's the same on each side). This would give (x+6)(x+2) = 4*4 x^2 + 8x + 12 = 16 x^2 + 8x - 4 = 0 (-8+or-sqrt(64-4*-4))/2 = x (-8+or-sqrt(80))/2 = x (-8+4*sqrt(5))/2 = x (I ditched the spurious negative result here) -4+2*sqrt(5) = x As x, the extra part of the new chord length, is 2*sqrt(5) - 4, and the remainder of that half chord to the centre is 4, the radius is 2*sqrt(5). It's a different way and a bit messier than yours, but might suit a small minority of students.
Superimpose the diagram with the reflection of the diagram about the line through the center with slope minus 1. The result is 2 rectangles, one which is 4 by 2 and one which is 2 by 4, sharing a corner, where the edges form straight lines. The center of the circle is halfway between the two other corners which are inside the circle. The distance from this point to point B is the radius, but this forms a right triangle which has height half of 4 and width 2 plus half of 4, or 2 and 4 respectively. Pythagorean theorem immediately gives root(4+16)=2*root(5). Even more intuitively, mirror everything a second time about the line through the center with slope 1. Then you have a plus shape which makes the calculation jump out even clearer.
QUITE CORRECT, BUT I BELIEVE YOU SHOULD USE FIRST THALEM' S THEOREM USING 3 POINTS ON CIRCLE AND THE RIGHT ANGLE, IN ORDER TO ESTABLISH THAT O IS THE CENTER BEFORE STEP 3, SO THERE IS SYMMETRY, CD=PF
Thanks Nicos for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I really enjoy watching your Videos because they are nice little riddles for our daily routine. You explain all this stuff without using really big and complicated math stuff and that is why I love Your videos
Please let me suggest: find length AC (we had a similar problem in one of your videos). Find sin(ABC). Use formula sin(a+b) ... Then AC/sin(ABC) is equel to 2R according to sinus theorm.
Chord theorem: the orthogonal intersection of two chord halves is the center of the circle. origin of the coordinate system = B Normal vector of chord 1 (6; 2) the midpoint of the chord 1 (3; 1) the straight line equation of the chord bisector is 6x + 2y = 20 Normal vector of chord 2 (0; -4) the midpoint of the chord 2 (0; -2) the straight line equation of the string bisector is y = -2 the intersection of the two lines (= center of the circle) 6x +2 (-2) = 20 => x = 4; y = -2 Pithagoras theorem radius = sqrt (16 + 4) = 4,472
Your approach is really nice. My approach also took same time.may be It depends how we train our Brain.I joined one end of the chord with length 4 to the other end of the lie segment of length 2 both away from the line segment of length 6. So the derived line segment would be a chord of the circle and will intersect the line segment of length 6. Then I Applied similar triangle. Which helped me in finding the two parts of line segment with length 6. Then I extended the line segment of length segment to complete a chord. Then joined the center and end point bof the chord. Then the missing part would be found also the perpendicular distance between the chord( when we extednd the line segment of length 6 it will give a length of the chord as 8hence half length will 4. This is statement is already derived using similar triangle)will be 2 hence using Pythagoras where length of perpendicular is 2 and base is 4. This will give a radius of sqrt(16+4) which will be 2sqrt(5)
For those asking why CD = PF Let's call the center of CF "S" CS=FS The radius is a perpendicular bisector of CF so it passes through S Also the radius bisects DP since DP // AB So DS=PS=2 We have CS=FS CD+DS=PF+PS CD+2=PF+2 (-2) CD=PF
5:42 why does AE contain O? I got the same answer in a different way, using law of cosines and inscribed angle conjecture. Let AC intersect BD at point Z. Then, ABZ and CDZ are similar triangles by AA. Easy to see the coefficient of similarity is 2; so BZ=4, DZ=2; using Pythagorean theorem, AC=AZ+CZ=6√2, BC=2√(10). Using law of cosines, cos
I used algebra and equation for circle to find intersecting points of a pair of circles giving me a line through the centre. Essentially what you do with a compass to construct a line passing through the centre of the circle. Doing this twice gives me intersecting lines passing through the centre (x,y) Now I can use the centre point and any other known point on the circumference of the circle to calculate my radius If you’re going to use Thales’ Theorem then you can’t just quote it - surely You have to demonstrate/build the proof as part of your answer
I'm using A,BC,D as described above On a coordinate plane, put A as (0,0), B as (0,4), D as (6, 4), and C as (6,6). Find the slope of BC, and the midpoint of BC (which is a chord). use that slope and midpoint to create and equation of its right bisector (y = -3x+14) Another chord AB has midpoint (0,2) and slope undefined. The equation of its right bisector is y = 2 find the point of intersection (POI) of the two right bisectors. (it's (4,2)). POI of the 2 Right Bisectors = center of circle. distance from (4, 2) to either of A, B, or C is 2*sqrt(5)
Another commenter pointed out that because of symmetry due to the inscribed rectangle What inscribed rectangle?, I asked myself. What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
From inscribed quadrilateral ABCF: ∠B+∠F=π (opposite angles); from AB∥FC and transversal BC: ∠B+∠C=π; hence ∠C=∠F, and ⊿BCD=⊿AFP (right triangles with BD=AP).
Saying “if A, B and C are points on a circle where line AB is a diameter.. then ACB is a right triangle” does NOT imply that any inscribed right triangle must have a diameter as its hypotenuse. It might be true (actually, it is true), but it is NOT a logical conclusion from Thales’ Theorum as stated. Just sayin’...
I played around with a piece of paper, and it looks like the geometry does go both ways. I think it would be less annoying if the statement was "A, B, and C are points on a circle with a line AB is a diameter of the circle if and only if the angle ACB is a right angle". Just a nitpick in the logic, that's all.
@@italixgaming915 well what don't you make a video and show the works how amazing you are. If you've got multiple solutions, this video probably isn't designed with you in mind. I learned a lot, for which in grateful.
We extend line BC to intersect the circle at a point E If we draw line AC it intersects line BD at a point G Comparing the similar triangles ABG and DCG BG is 4 units and GD 2 units. (BG/GD=AB/CD=2 units) So, in the triangle CDG, DG=DC => angle GCD=angle CGD=45° => angle BAC=45° => angle BEC=45° (same arc BC) =>DE=DC=2 units =>BE=8 units => in triangle ABE AB^2+BE^2=AE^2 =>AE^2=16+64=80 angle ABE=90° => AE is diameter so r=√80 /2
Extend BD to the right to intersect with the circle at E. Extend CD down to intersect with the circle at F. Let G be a point on DF where AG is perpendicular to DF. As BA is a chord, B and A are equidistant from O, and since CF is also a chord and DB and AG are perpendicular to CF, D and G are also equidistant from O and AGDB is a rectangle and vertically symmetrical about O. By this symmetry, DB = AG = 6, CD = GF = 2, and BA = DG = 4. As chords CF and BE intersect, the products of their respective segments on either side of D will be equal: BD(DE) = CD(DF) 6x = 2(4+2) = 12 x = 12/6 = 2 Draw a line from A to E. As ∠EBA is a right angle and E, B, and A are all on the circumference of circle O, AE is a diameter and passes through O. Triangle ∆EBA: AB² + BE² = EA² 4² + (6+2)² = (2r)² 16 + 64 = 4r² r² = 80/4 = 20 r = √20 = 2√5
I love the fact that there is often more than one way to solve geometrical problems. Taking B as the origin, I found the equations of the perpendicular bisectors of AB and BC. I then determined the point of intersection, which is the centre of the circle. Then I used Pythagoras Theorem to find the radius length.
@@joshwalker5605 it's true, but i need a mathematical proof for it. Rectangle and circle are symmetrical so when they're aligned like an outcircle parallel chords will be cut equally. Would be interesting to see an elaborate mathematical proof.
BA=4; DB=6; DC=2; if we extend BD and CD, they cut the circumference at points E and F respectively. By symmetry, the perpendicular bisector of BA and that of CF is the same and determines a diameter that, taken as the axis of symmetry, allows us to deduce that CF=CD+BA+DC=2+4+2=8 Power of point D with respect to the circumference: CDxDF=BDxDE → 2(8-2)=6DE → 2x6=6DE → DE=2 → BE=BD+DE=6+2=8 Triangle ABE is rectangular and is inscribed in the circumference, from which it follows that its hypotenuse is a diameter, that is, AE=2r → AB²+BE²=AE²=(2r)² → 4²+8²=4r²→ r²=80 /4=20 → r=√20=2√5
So nice of you Will! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
I really enjoy your challenging questions. Using your method the reader must know a lot of geometry. Is this deliberate? There is a much simpler way just using similar triangles and the fact that a perpendicular through the midpoint of a chord passes through the circle center. This is the method used when first learning technical drawing at school.
Great, did this myself, but I was lucky because I extended the 6 unit line to be a cord just as you did but I didn't prove it as you did, I assumed it was 2 extra units, so i was luck, oh and btw I didn't know about Thales theorem, so that's even more you've taught me, thanks as always
1) On a Cartesian coordinate, we would have A(0,-2), B(0,2), C(6,4) and O should be (x,0). 2) Since O is the center of the circle, we have OB = OC or x^2 + 2^2 = (x-6)^2 + 4^2, we then have x = 4 3) The radius is OB^2 = (4-0)^2 + (0-2)^2. OB = 2*sqrt(5)
I did this one in my mind. I used a slightly different method, and got the correct answer. I dropped perpendiculars from O onto the mid points of AB and CF. So they'll pass through the centre. Since AB and CF are parallel, the perpendiculars are basically the same line. Then you just mark one part as a and the other as 6-a (This perpendicular is parallel and same distance as BD). a^2 + 2^2 = r^2. This is because 2 is half of AB (mid point). Do the same with 6-a and 4 (half of 8). Equate to get a. Then from that you'll get r. It's easy to do in the mind then.
The game is to do them on mind :)) My rules: you can see only first screen, you cannot touch the screen or anything. Often I can, sometimes I cannot. But it's funny. I used a different method. From A, segment parallel to BD, lenght 2. From that, a segment parallel to AB. This has to touch the circle (in Z, we say) when its lenght is 6. finally segment from Z parallel to BD: it reaches C after a lenght of 4. Flip all over axis of AB. You has a rectangle 4 x 8, the diagonal is the diameter, math is easy.
Very easy with analytical geometry: A is (0,0), the center is (x0,y0), then x0^2 + y0^2 = r^2 (point A) x0^2 + (4 -y0)^2 = r^2 (point B) (6 - x0)^2 + (6 - y0)^2 = r^2 (p. C) From eq. 1 & 2 we obtain y0 = 2 then from eq. 1 & 3, and y0 = 2 we obtain x0 = 4 Then eq. 1 is 2^2 + 4^2 = r^2
Nice! But at 05:00 it's practically done, as there is another cord theorem stating that four times the square of the radius is equal to the sum of the squares of the segments of two perpendicular cords. That is, 4r² = a² + b² + c² + d². So, 4r² = 6² + 2² + 2² + 6² = 80².
@@alin1553 I don't think it has a special name. It is just a theorem arising from the Power of a Point Theorem. It can be proved algebraically (most common) or geometrically.
Another commenter pointed out that because of symmetry due to the inscribed rectangle What inscribed rectangle?, I asked myself. What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
We can imagine that the side that has a length of 2 units can be rotated by 90 degrees. We can then form a side length of 6+2=8. This would then form the diameter of the circle , which we could find thru Pythagoras Theorem: (4)^2 + (8)^2 = Root(80)=8.944. But we're not done, radius = 4.472 units -> pi×r^2 = 62.828 Therefore, the Area of the circle is ~62.828
I would set an axis system in point A. Then I'll get 3 points: A(0,0), B(0, 4) and C (6, 6). then I can caluclate radius by using equation of the circle. (x-a)^2 + (y-b)^2 = r^2 where r is radius of circle and a,b are cordinates of the centre of the circle. I have 3 unknow values and 3 points.
Hello, Would you please explain how you determine that a line from the AB-line center crosses the diameter of the circle perpendicularly; or O, the center of the circle?
@@Abhishek-hk1on Hello, Forget what idiots claim and their theories; everything has theories. I can prove "that circle diameter makes 4.0e-¹°, 2.0e¹°, or 4.800e³°, ETC." People or mathematicians could fool around with mathematics; nevertheless, they could do it with the universe's laws; because it is God's order, and he (God) respects and follows it. God says, "Before the creation, I (God) created logic (physic)," and every matter or manner must follow it. I can also prove that this life is not |absolute| and is an experience of the past-future for the soul of creatures. Studying nuclear cosmology and physic make these phenomena easier to comprehend. Your question or reply doesn't satisfy my question; however, I thank you so much. Have a beautiful weekend.
*Much Simpler Solution* - Drop the 2-units line down by 4 units (completing the 6x4 rectangle) and then again by symmetry 2 more units to touch the circle, thereby changing this into a new problem of determining the circum-radius of triangle of sides (a=2+4+2=8), altitude h = 6, b^2=36+36 and c^2 = 36+4. Area = ah/2 = 8x6/2 = 24. abc = (8)(6√2)(2√10) = 192√5. Circum-Radius R = abc / (4 x Area) = 192√5 / (4x24) = 2√5. *Simple*
Draw the line AC crossing BD at F. Then FBA is similar to FDC, and therefore AB/CD = BF/FD, therefore BF = 4, FD = 2, therefore all the acute angles of these two triangles = pi/4. Extend BD to meet the circle at E, then ACE is another right-angle triangle (hypotenuse passes through the origin), the angles at C add to pi/2, therefore DCE has acute angles = pi/4. Therefore DE = CD = 2, and BE = 6 + 2 = 8 ==> r = sqrt(20).
Connect A to C and you have two similar triangles and it is easy to find that they are also Isosceles right triangles. Angle ACD is 45°, so the angle DCE is also 45° and DE is 2 ...
Draw line AC. Notice that this is the diagonal of a 6x6 square, so angle BAC is 45°. This is an angle subtended from chord BC to the circumference, so angle BOC is 45° × 2 = 90°. Therefore 2r^2 = BC^2 = 6^2 + 2^2 = 40, so r = √20 = 2√5.
AB is a chord. So line passing through midpoint of AB will be a diameter. This diameter cuts PD at its midpoint. Since this a circle, by symmetry CD=PF proved...
At around the three minute mark, You say that CD=PF; how do we know this? How do I know to apply symmetry to this diagram? Am I overlooking some simple logic? Please, and thank You. B
Symmetry. Inscribe any square or rectangle and any line from it, square to the line of it, to the circle, will be the same on the opposite side of the square/rectangle (He just didn't complete the rectangle - extend AP to the circle on the right, draw vertical from that point to E and the inscribed rectangle is completed)
@M Simpson1) I said "inscribed square or rectangle" Going by picture at 1:57 AB and BE are two lines of an inscribed rectangle. Extend AP to the right so it reaches the circle and you have the 3rd line (parallel to BE). From this new unnamed point, draw a vertical line to E and that's your 4th line, parallel to AB. And there's your complete inscribed rectangle. Symmetry axis is a horizontal line (parallel to BE and AP) through O
I solved it in my head, by noticing the angle BAC is 45 degrees. (B is directly up, while C is 6 units up and 6 units right.) Because the angle BAC is 45 degrees, the arc BC is 90 degrees (double the angle). So BOC is a 45-45-90 right triangle. The length of BC is sqrt(6*6+2*2) = sqrt(40) Thus, the radius (OC or OB) is sqrt(40)/sqrt(2) = sqrt(20) = sqrt(4*5) = 2*sqrt(5).
I solved it in one line in my mind: Area of triangle ABC is equal to S=12 (side = 4, height = 6). Another simple formula for area of triangle is S = abc/(4R). Or 12 = abc/(4R) or R = abc/48, where a, b,c are the sides of triangle. Sides of the triangle ABC are following: 4, 6sqrt(2) and 2sqrt(10). After multiplication of sides we get R = 2sqrt(5).
So, I have one question. If you only have whole numbers and zero significant digits, how did you get to three significant digits accuracy for your answer?
How can we assume that a chord starting from A, extending through O, will land at point E? Is there some symmetry or rule of circles I’m missing? I mean this diagram isn’t going to be to scale?!
Focus on triangle ABC: Let the 3 sides of ABC be a, b, c, It's easy to get a = BC = 2√10, b=AC=6√2, c=AB=4. From the theorem of cosine: cos B = (a^2 + c^2 - b^2) / 2ac = -1/√10 (since we already know a, b, c). This implies, sin B = 3/√10. Support the R is the radius of the circumscribed circle of triangle ABC. From the theorem of Sine: we have a/sinA = b/sinB = c/sinC = 2R. From 2R = b/sinB, we have R = 2√5. And, the circle in the picture is just the circumscribed circle of triangle ABC. Thus the answer is 2√5.
Alternatively you can see that 2+4 =6 and build back the figure from the circle center : Take a point O on a grid from O you can build A by going two step down and four step left from O you can build B by going two step up and four step left from O you can build C by going two step right and four steps up all those path form congruent right angle triangles of sides 2 and 4, hence O is at a distance 2sqrt(5) of A,B and C : we have the radius and the center.
Analytical solution: define point coordinates A(0, -2), B(0, 2), D(6, 2), C(6, 4). O (circle center) will be at (x, 0) (x axis is perpenticular to AB and has origin at intersection with AB, y axis is AB with y=0 at the middle of AB. O is at the intersection of medians of AB and BC. BC median has equation y=-3x + 12 (BC center is at (3,3)). Solve for y = 0 yields x=4. Then radius is the distance OA (or OB or OC, it does not matter since they're all the same), sqrt(4^2 + 2^2) = 2*sqrt(5).
People don't realize that since A,B,D is a rectangle, it doesn't matter how big it is. Think about it. Because point A and B touch the circle wall, it's impossible for there to be different gaps above and below it. C,D and P,F will always be the same as long as it's a rectangle and two points touch the wall. As such, no matter how big or small the rectangle is, A and E will always be the diameter passing through the center.
Another way of solving the problem: . Overlay the circle with a coordinate plane where the left chord has endpoints (0,0) and (0, -4). The perpendicular bisector of this chord has the formulas y=-2 and runs through the center of the circle. Draw the hypotenuse of the right triangle with sides of length 6 and 2. The coordinates of the endpoints of this chord are (0,0) and (6, 2). Its midpoint is (3, 1), and its slope is 1/3. Its perpendicular bisector is y=-3x+10, and it also runs through the center of the circle. The two perpendicular bisectors intersect at the point (4, -2), which the center of the circle and is a distance of 2*sqrt(5) from (0,0).
How can you prove CD = PF = 2? Just because you made a rectangle in the middle doesnt mean you can say "symmetry" and move the other distance to the bottom at 3:20
This proof relies not on Thale's Theorem as stated, but on its converse: the center of the circumcircle of a right triangle lies on its hypotenuse. I think the video would be more helpful if the basic approach of the proof had been outlined up front, thus motivating the steps.
Are we assuming no access to a ruler or tape measure? Because the radius of a circle is half the diameter, so if you know the diameter just divide by 2 and that is the radius or am I just missing something?
Wow this is great... nevertheless I have another interesting solution to this.... ...... we know perpendicular bisector of a chord passes through centre 'o' of a circle. Lets coordinate the point B as (0,0) and so C is (6,2) So perpendicular bisector of AB chord is y = - 2 line.. Slope of BC line = 1/3 => slope of line perpendicular to BC is -3 Midpoint of BC is (6/2,2/2) or (3,1) so equation of the line through midpoint of BC and perpendicular to it is 3x+y-10=0 Intersection of the lines y=-2 and 3x+y-10=0 is the centre which by solving is O(4,-2) .....note : B is (0,0) .... Radius= Distance BO = 2✓5 .....thats it
I approximately almost got the answer just by looking at it though, I am curious of the perfect solution of the problem. I really love this kind of videos. Keep it coming 😊😊😊
I have another solution for mathematicians: For this kind of a problem when you have intersecting chords, and the sides are a, b, c, d just like you named them, you can. ALWAYS find the radius of the circle by this formula : 4r^2 = a^2 + b^2 + c^2 +d^2 Where r is the radius of the circle. Hint: ^2 means squared e. g r^2 means r squared. You can thank me later!
It's as simple using the coordonates of the point, the origin being B: B(0,0), A(0, -4), C(6,2), and using the circle equation (x-a)^2 + (y-b)^2= r. R being the radius and a,b the coordinates of the center. 1) B being on the circle: a^2 + b^2= r^2 2) A being on the circle: a^2 + (b+4)^2= r^2. Wich gives by substracting 1) b=-2 and a^2= r^2 - 4 3) C being on the circle: (6-a)^2 + (2-b)^2= r^2 ==> (6-a)^2 + 16= a^2+4 from 2). So (6-a)^2= a^2-12, i.e. 36 + a^2 - 12a= a^2 -12 ==> a=4 4) r^2= 16+4=20 so r= sqrt(20)= 2*sqrt(5).
Much simpler solution... if you draw AC, it works out that you have two similar triangles, and that angle BAC is 45 degrees. BC is therefore the chord of a 90 degree radial angle, length 2 root 10. That leads to a radius of 2 root 5.
Consider the points A,B,C,are points on the circle. Let O be the centre say (x,y). Co ordinates of A(0,0)B(0,4) and C(6,6) Solve for OA=OB,we get y=2 Solve for OB=OC we get x=4 Hence radius= ✓4^2+2^2=✓20=2✓5 is correct
O lower than B at 2. Middle of BC higher at 1 and righter at 3 fron B. O lower than that middle at 3. Right shift of O 2/6 of that high and equal 1. Takes O lower at 2 and righter at 4. So radius is √(2^2+(3+1)^2)=√20=2√5 Seems thais way a bit shorter... Sorry if my english poor - learn it at other century :) and almost don't use in life.
i love watching this guy.. i used to have to work things like this out on paper years ago as a mechanical precision , but then i worked for other companies where they have auto cad. so i didn't need to work it out anymore, because the guys gave me the answers,.. well done teacher.. i love watching your videos.
So nice of you Paul! You are awesome 👍 Thank you so much for taking the time to leave this comment. I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
@@italixgaming915 HAHA I CALL THIS BRAINLESS SOLUTION (NO OFFENSIVE
I am also lazy in my old age, I can do this the way shown but I just timed myself using CAD - drawing the three lines, drawing a circle from three points and finding the diameter of the circle. It took 90 seconds and that included starting the CAD program up. It's just another tool like a pen and paper, slide rule or calculator. We all take the easy way out if we can.
I find his solution significantly less complicated than yours, better explained and more polite.
@@xoxb2 yeah atleast he isn't bragging about how much time it took reducing all the steps he took to make idiots like that guy understand he would probably solve in half his time
Put A to coordinate (0,0), B to (0,4) and C to (6,6).
Put the coordinates into the equation of the circle. You get 3 unknown (coordinates for circle center and r) and 3 equations. Solve for (x,y) för center of circle and radius. If you have the coordinates for 3 points on the circle you can draw the circle. More algebra, less geometry.
I did it this way ;)
Same me
Yeah. He’s made this way harder than it needs to be.
@@danarrington2224 it's not hard, just time consuming xd
Even easier: Let the circle center be (0,0), then A =(x,-2) and B=(x,2) because of symmetry. C=(x+6,4). |C|^2=r^2=|B|^2, which reduces to a linear equation in x.
Without any geometry, you can find the radius using coordinates and a little algebra. Put the origin at the point A (0,0), then B is (0,4) and C is (6,6).
You know the equation of the circle is of the form (x+a)^2 + (y+b)^2 = r^2
Substituting x and y from the three points gives three equations:
a^2 + b^2 = r^2
a^2 + b^2 + 8b + 16 = r^2
a^2 + b^2 + 12a + 12b + 36 +36 = r^2
You can subtract the first equation from each of the second and third equations giving:
8b + 16 = 0
12a +12b +72 = 0
The former gives:
b = -2
and then using that in the latter gives:
12a - 24 + 72 = 0
which leads to a = -4
Since r^2 = a^2 + b^2, we get:
r^2 = 16 + 4 = 20
so r = sqrt(20) or 2.sqrt(5) ~= 4.472
No constructions; no "by symmetry"; no intersecting chords; no angle in a semicircle; no Pythagoras, just a pair of linear simultaneous equations.
My solution as well. Cut to the crap. But as an old math teacher it is nice to see the connection btw pythagoras, Thales, corda. I would show both methods. If I still worked as a teacher.
I do same way. But, the equation of the circle based on pythagoras theorem, so this method use this.
I did something similar, except I didn't move the origin, I only rotated the circle until the cord ab was perfectly vertical. so my points were a(x,-2) b(x,2) and c(x+6, 4)
substituting a and c into the equation we get x^2 + 4 =x^2 + 12x +52
solve for x => 4 = 12x + 52 => 12x = 4-52 => x = -48/12 = -4
we plug our variables back into the circle equation and solve for r, we get r = sqrt(20) = 2sqrt(5)
Another solution.
Let OA=OB=OC=r.
Draw a perpendicular line from point O to line AB, and let the intersection point be E.
Let the length of line EO be x.
Since triangle OAB is an isosceles triangle
(a) r^2 = 2^2 + x^2
Extend line CD and line EO, and set the intersection point to F.
Since the length of the line segment CF will be 4
(b) r^2 = 4^2 + (6-x)^2
From (a) and (b)
2^2 + x^2 = 4^2 + (6-x)^2
Therefore x=4
Substituting x=4 into (a), we get r=2 Sqrt (5)
I used this method too
why will CF be 4?
@@bono95zg Becaus it is CF = CD + half of AB = 2 + 4/2 = 2 + 2 = 4.
Smart! !
I've got to go back to school to learn about Thales' and Chord theorems. Never heard of them. Thanks.
me too Hhahaha i was like , "Chords Theorems?" . I know what were Chords but not their theorems
That is because in school you dont see that theorems, neither other much theorems
If you went to school and never heard of them what makes you think going back would help? If you heard about them on your own why do not seek self study instead?
@@BigLeagueDrew it's just an expression
Lol also never heard this
PF = CD was not adequately explained.
Indeed. Whenever somebody says "by symmetry", the obvious response should always be: "Prove it!"
Right, he just pulled symmetry out of the air as far as I can tell.
Another commenter pointed out that because of symmetry due to the inscribed rectangle
What inscribed rectangle?, I asked myself.
What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
Thanks, I finally got it when I realized that one edge of the square or rectangle had to be a chord. Seems obvious now. Love math, subscribed! And please pardon the double post. :)
Yes
(1) 3:01 Symmetry was assumed, not proven.
(2) 6:38 The _converse_ of Thales theorem was used.
True, these inaccuracies are very annoying.
Did he demonstrate that the AE segment is a diameter?
Once you know all 4 chord segments, you can use 4r²=CD² + DF² + BD² + DE². The reason why the Pythagorean theorem works in this case is an inscribed angle in a circle is 1/2 its intercepted arc. Since angle ABE is 90 degrees arc AFE is 180 degrees. When the line is drawn from A to E, it must pass through the center of the circle.
Sir, what is the "Symmetry property" based on which you have used CD=PF?
thats what i am saying. XD itt made no sense to me
Yes. It looks intuitively correct, but how is he proving it?
same question here, visually it is correct, I even draw it at scale to make sure, but I don't have a proof
The Chords Theorem was new for me! I have looked it up on Wikipedia, that makes trigonometry in circles so much easier :D Thanks for the Video!
I solved it this way.
*Draw line segment AC and using similar triangles we can prove angle BAC is 45°
*Next mark center O and draw triangle BOC. Angle BOC is 90° using arc-centre degree measure theorem
*BC can be computed by applying Pythagorean theorem on BDC
*Now in right triangle BOC, apply Pythagorean theorem to find out the radius of the circle
But BAC is not 45
I think you applied the converse of Thales' Theorem: If A, B, and C are points on a circle where the angle ACB is a right angle, then the line AB is a diameter of the circle. Of course, this has also been proven true.
Exactly. That's what I came here to say. The fact that A --> B does NOT by itself mean that B --> A. In fact, simply claiming that it does is a classic logical fallacy (affirming the consequent.) So this was not a proper application of Thales' Theorem. Now, it appears that the converse of Thales' Theorem is in fact ALSO true, but it should be treated as a separate theorem, to make it clear that the solution of the problem here is not proceeding illogically. It needs to be stated categorically that B --> A independently.
It's because of the circumscribed circle, not the Thales' Theorem.
Also came here to say this!
Well said, this is the part not clear in the video
@@danielvieira8374 Because he did not explain why the center of the circle is a point on the segment.
Nice, I've got the right answer without knowing any of those theorems. I've just calculated the radius using phythagoras with r = √(2^2+4^2). With 2 beeing half the lenghth of the left line (symmetry) and 4 beeing the lenghth of the horizontal from the point on the circle to the intersection of the horizontal line and the conecting line from the point on the circle on the bottom and the the point on the circle on the top.
If you take Pythagoras theorem from the bottom of 4 to the top of 2, which is what I think you're doing, wouldn't you get √((4^2+4^2)+(2^2+2^2))?
you haven't actually pointed out how that line goes through the center of the circle though, so it can't be used to determine the radius.
If you draw a parallel to BE that goes through A, then CF is divided into three parts that must - according to symmetry - have the length 2 - 4 - 2. DF therefore has a length of 6 and is therefore the same length as BD. Therefore CD is equal to DE. So we have a rectangle inside the circle with the corners A, B, E and the second intersection of the drawn parallel with the circle. The rectangle has side lengths 8 and 4.
Thus, the diagonal of the rectangle is d, and we have:
d² = 4² + 8² = 16 + 64 = 80
r = 1/2 * d and thus r² = 80 / 4 = 20
r is therefore 2 * sqrt(5) = 4.472 long.
My method was quite different.
You have the cord AB and can make a second cord BC. We know that a perpendicular line from the mid point of a cord passes through the centre of the circle.
The line BC has a slope 2/6 =1/3 relative to the horizontal.
The centre of BC will be 3 to the right from B and 1 above B.
The centre of the circle will be 3 below this point.
The slope of the perpendicular line from BC 1/3 relative to the vertical.
The centre of the circle will therefore be 1 to right of the centre of BC.
The centre will be 3+1 =4 to the right of AB.
Half of AB = 2.
Pythagoras time.
R^2 = 4^2 + 2^2 = 16 + 4 =20
R = root 20 =2 root 5
Indeed. The center of the circle will be on a line parallel to and 2 below of AD; that line is perpendicular to AB and cuts it in half. It will also be on the perpendicular line that cuts AC in half; it cuts that line 1 above AD. As AC is sloped 3/1 it is sloped 1/3, and as the center is 3 lower it is one to the right of the middle of AC, so 3+1 = 4 to the right of AB. Then r = √(4² + 2²) = √20 = 2√5.
That's much cleaner than how I often do it. Once I have calculated chord lengths, I might move a chord to become the diameter.
For instance, moving BE down to cross the centre would have it as (x+6)(x+2) where x is the distance between chord end and circumference (it's the same on each side). This would give (x+6)(x+2) = 4*4
x^2 + 8x + 12 = 16
x^2 + 8x - 4 = 0
(-8+or-sqrt(64-4*-4))/2 = x
(-8+or-sqrt(80))/2 = x
(-8+4*sqrt(5))/2 = x (I ditched the spurious negative result here)
-4+2*sqrt(5) = x
As x, the extra part of the new chord length, is 2*sqrt(5) - 4, and the remainder of that half chord to the centre is 4, the radius is 2*sqrt(5).
It's a different way and a bit messier than yours, but might suit a small minority of students.
My method was similar. I calculated the missing parts of the intersecting chords and then used Pythagoras to calculate the radius directly.
I agree. There are simpler ways.
Superimpose the diagram with the reflection of the diagram about the line through the center with slope minus 1. The result is 2 rectangles, one which is 4 by 2 and one which is 2 by 4, sharing a corner, where the edges form straight lines. The center of the circle is halfway between the two other corners which are inside the circle. The distance from this point to point B is the radius, but this forms a right triangle which has height half of 4 and width 2 plus half of 4, or 2 and 4 respectively. Pythagorean theorem immediately gives root(4+16)=2*root(5). Even more intuitively, mirror everything a second time about the line through the center with slope 1. Then you have a plus shape which makes the calculation jump out even clearer.
Thanks, I finally got it when I realized that one edge of the square or rectangle had to be a chord. Seems obvious now. Love math, subscribed!
QUITE CORRECT, BUT I BELIEVE YOU SHOULD USE FIRST THALEM' S THEOREM USING 3 POINTS ON CIRCLE AND THE RIGHT ANGLE, IN ORDER TO ESTABLISH THAT O IS THE CENTER BEFORE STEP 3, SO THERE IS SYMMETRY, CD=PF
Thanks Nicos for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
@@PreMath That's not what he meant... prove it....
I really enjoy watching your Videos because they are nice little riddles for our daily routine.
You explain all this stuff without using really big and complicated math stuff and that is why I love Your videos
Pls explain further about symmetry property.Tks.
Please let me suggest: find length AC (we had a similar problem in one of your videos). Find sin(ABC). Use formula sin(a+b) ... Then AC/sin(ABC) is equel to 2R according to sinus theorm.
Wow! You, sir, are like the Sherlock Holmes of maths! Totally engrossing, as always. Thank you!
Chord theorem: the orthogonal intersection of two chord halves is the center of the circle.
origin of the coordinate system = B
Normal vector of chord 1 (6; 2)
the midpoint of the chord 1 (3; 1)
the straight line equation of the chord bisector is 6x + 2y = 20
Normal vector of chord 2 (0; -4)
the midpoint of the chord 2 (0; -2)
the straight line equation of the string bisector is y = -2
the intersection of the two lines (= center of the circle)
6x +2 (-2) = 20 => x = 4; y = -2
Pithagoras theorem
radius = sqrt (16 + 4) = 4,472
Not understood sir can you make a video plz
And send it on this
comment
Thanks. I didn't know what *Intersecting Chords Theorum* is till date. Thanks for the same, now I know.
Your approach is really nice. My approach also took same time.may be It depends how we train our Brain.I joined one end of the chord with length 4 to the other end of the lie segment of length 2 both away from the line segment of length 6. So the derived line segment would be a chord of the circle and will intersect the line segment of length 6. Then I Applied similar triangle. Which helped me in finding the two parts of line segment with length 6. Then I extended the line segment of length segment to complete a chord. Then joined the center and end point bof the chord. Then the missing part would be found also the perpendicular distance between the chord( when we extednd the line segment of length 6 it will give a length of the chord as 8hence half length will 4. This is statement is already derived using similar triangle)will be 2 hence using Pythagoras where length of perpendicular is 2 and base is 4. This will give a radius of sqrt(16+4) which will be 2sqrt(5)
For those asking why CD = PF
Let's call the center of CF "S"
CS=FS
The radius is a perpendicular bisector of CF so it passes through S
Also the radius bisects DP since DP // AB
So DS=PS=2
We have CS=FS
CD+DS=PF+PS
CD+2=PF+2 (-2)
CD=PF
5:42 why does AE contain O?
I got the same answer in a different way, using law of cosines and inscribed angle conjecture.
Let AC intersect BD at point Z. Then, ABZ and CDZ are similar triangles by AA.
Easy to see the coefficient of similarity is 2; so BZ=4, DZ=2; using Pythagorean theorem, AC=AZ+CZ=6√2, BC=2√(10).
Using law of cosines, cos
Yeah, he just assumed the center O was a point on AE...did not show or prove why it is.
@@DL101ca YES!!! That was crazy how he assumed that. Very weird and frustrating.
I used algebra and equation for circle to find intersecting points of a pair of circles giving me a line through the centre. Essentially what you do with a compass to construct a line passing through the centre of the circle. Doing this twice gives me intersecting lines passing through the centre (x,y)
Now I can use the centre point and any other known point on the circumference of the circle to calculate my radius
If you’re going to use Thales’ Theorem then you can’t just quote it - surely
You have to demonstrate/build the proof as part of your answer
I'm using A,BC,D as described above
On a coordinate plane, put A as (0,0), B as (0,4), D as (6, 4), and C as (6,6).
Find the slope of BC, and the midpoint of BC (which is a chord). use that slope and midpoint to create and equation of its right bisector (y = -3x+14)
Another chord AB has midpoint (0,2) and slope undefined. The equation of its right bisector is y = 2
find the point of intersection (POI) of the two right bisectors. (it's (4,2)). POI of the 2 Right Bisectors = center of circle.
distance from (4, 2) to either of A, B, or C is 2*sqrt(5)
Hello, please explain, why do you consider that CD = PF?
Another commenter pointed out that because of symmetry due to the inscribed rectangle
What inscribed rectangle?, I asked myself.
What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
From inscribed quadrilateral ABCF: ∠B+∠F=π (opposite angles);
from AB∥FC and transversal BC: ∠B+∠C=π;
hence ∠C=∠F, and ⊿BCD=⊿AFP (right triangles with BD=AP).
Saying “if A, B and C are points on a circle where line AB is a diameter.. then ACB is a right triangle” does NOT imply that any inscribed right triangle must have a diameter as its hypotenuse. It might be true (actually, it is true), but it is NOT a logical conclusion from Thales’ Theorum as stated. Just sayin’...
That is what I thought. That was agrivating the crap out of me.
Presumably it is another theorem. Let's call it the Phillips theorem ;)
Thanks - you saved me writing a comment! Or, wait a minute... I just did anyway 🤣
I played around with a piece of paper, and it looks like the geometry does go both ways. I think it would be less annoying if the statement was "A, B, and C are points on a circle with a line AB is a diameter of the circle if and only if the angle ACB is a right angle".
Just a nitpick in the logic, that's all.
@@jefftaylor8320 aggravating, too.
At a quick look, the center is at the intersect of two chords midpoints. One chord we have the length 4, the other we form from sqrt(36+ 4).
Awesome problem that brings lots of properties of circles together. Great work!
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
I'll visit your channel soon as well!
@@italixgaming915 well what don't you make a video and show the works how amazing you are.
If you've got multiple solutions, this video probably isn't designed with you in mind. I learned a lot, for which in grateful.
We extend line BC to intersect the circle at a point E
If we draw line AC it intersects line BD at a point G
Comparing the similar triangles ABG and DCG BG is 4 units and GD 2 units. (BG/GD=AB/CD=2 units)
So, in the triangle CDG, DG=DC
=> angle GCD=angle CGD=45°
=> angle BAC=45°
=> angle BEC=45° (same arc BC)
=>DE=DC=2 units
=>BE=8 units
=> in triangle ABE AB^2+BE^2=AE^2
=>AE^2=16+64=80
angle ABE=90° => AE is diameter so r=√80 /2
Extend BD to the right to intersect with the circle at E. Extend CD down to intersect with the circle at F. Let G be a point on DF where AG is perpendicular to DF. As BA is a chord, B and A are equidistant from O, and since CF is also a chord and DB and AG are perpendicular to CF, D and G are also equidistant from O and AGDB is a rectangle and vertically symmetrical about O. By this symmetry, DB = AG = 6, CD = GF = 2, and BA = DG = 4.
As chords CF and BE intersect, the products of their respective segments on either side of D will be equal:
BD(DE) = CD(DF)
6x = 2(4+2) = 12
x = 12/6 = 2
Draw a line from A to E. As ∠EBA is a right angle and E, B, and A are all on the circumference of circle O, AE is a diameter and passes through O.
Triangle ∆EBA:
AB² + BE² = EA²
4² + (6+2)² = (2r)²
16 + 64 = 4r²
r² = 80/4 = 20
r = √20 = 2√5
I love the fact that there is often more than one way to solve geometrical problems. Taking B as the origin, I found the equations of the perpendicular bisectors of AB and BC. I then determined the point of intersection, which is the centre of the circle. Then I used Pythagoras Theorem to find the radius length.
I just KNEW it would come to triangles. Just EVERYTHING in mathematics is about triangles. They're everywhere! 🥰
Beautifully solved professor! 👏👏👏😇
Where can i find elaboration for the symmetry property where CD = PF. Any video on it's proof?
yeah i am not convinced this is true.
@@joshwalker5605 it's true, but i need a mathematical proof for it. Rectangle and circle are symmetrical so when they're aligned like an outcircle parallel chords will be cut equally. Would be interesting to see an elaborate mathematical proof.
BA=4; DB=6; DC=2; if we extend BD and CD, they cut the circumference at points E and F respectively. By symmetry, the perpendicular bisector of BA and that of CF is the same and determines a diameter that, taken as the axis of symmetry, allows us to deduce that CF=CD+BA+DC=2+4+2=8
Power of point D with respect to the circumference: CDxDF=BDxDE → 2(8-2)=6DE → 2x6=6DE → DE=2 → BE=BD+DE=6+2=8
Triangle ABE is rectangular and is inscribed in the circumference, from which it follows that its hypotenuse is a diameter, that is, AE=2r → AB²+BE²=AE²=(2r)² → 4²+8²=4r²→ r²=80 /4=20 → r=√20=2√5
Thanks for the example using the Chords and Thales' Theorems. Nice to have those tools.
So nice of you Will! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Applying the theorem of getting the radius from a circumscribed triangle works as well. Just have to get BC and AC (easily). 2R = AC / sin（ABC ）
so what is angle ABC , & what is his sin ?
its not easier !
@@vladlevkovets7087 inverse sin of (BD/AC). With a calculator of course
@@Hamsters831 less steps , but impossible way with only paper & pen )))
I really enjoy your challenging questions. Using your method the reader must know a lot of geometry. Is this deliberate? There is a much simpler way just using similar triangles and the fact that a perpendicular through the midpoint of a chord passes through the circle center. This is the method used when first learning technical drawing at school.
Awesome video, nice solution, many thanks!
AB = a = 4 → BE = 2a → AE = a√5 → r = (a√5)/2 = 2√5 🙂
or:
4 + 36 + 4 + 36 = 80 = (4)r^2 → r = 2√5 🙂
Great, did this myself, but I was lucky because I extended the 6 unit line to be a cord just as you did but I didn't prove it as you did, I assumed it was 2 extra units, so i was luck, oh and btw I didn't know about Thales theorem, so that's even more you've taught me, thanks as always
1) On a Cartesian coordinate, we would have A(0,-2), B(0,2), C(6,4) and O should be (x,0).
2) Since O is the center of the circle, we have OB = OC or x^2 + 2^2 = (x-6)^2 + 4^2, we then have x = 4
3) The radius is OB^2 = (4-0)^2 + (0-2)^2. OB = 2*sqrt(5)
I did this one in my mind. I used a slightly different method, and got the correct answer. I dropped perpendiculars from O onto the mid points of AB and CF. So they'll pass through the centre. Since AB and CF are parallel, the perpendiculars are basically the same line. Then you just mark one part as a and the other as 6-a (This perpendicular is parallel and same distance as BD). a^2 + 2^2 = r^2. This is because 2 is half of AB (mid point). Do the same with 6-a and 4 (half of 8). Equate to get a. Then from that you'll get r. It's easy to do in the mind then.
The game is to do them on mind :)) My rules: you can see only first screen, you cannot touch the screen or anything. Often I can, sometimes I cannot. But it's funny.
I used a different method. From A, segment parallel to BD, lenght 2. From that, a segment parallel to AB. This has to touch the circle (in Z, we say) when its lenght is 6. finally segment from Z parallel to BD: it reaches C after a lenght of 4. Flip all over axis of AB. You has a rectangle 4 x 8, the diagonal is the diameter, math is easy.
Geometry is one part of mathematics that seems much better taught in India than the West.
Very easy with analytical geometry:
A is (0,0), the center is (x0,y0), then
x0^2 + y0^2 = r^2 (point A)
x0^2 + (4 -y0)^2 = r^2 (point B)
(6 - x0)^2 + (6 - y0)^2 = r^2 (p. C)
From eq. 1 & 2 we obtain y0 = 2
then from eq. 1 & 3, and y0 = 2
we obtain x0 = 4
Then eq. 1 is
2^2 + 4^2 = r^2
Nice! But at 05:00 it's practically done, as there is another cord theorem stating that four times the square of the radius is equal to the sum of the squares of the segments of two perpendicular cords. That is, 4r² = a² + b² + c² + d². So, 4r² = 6² + 2² + 2² + 6² = 80².
Please let me know what's the name of this theorem?
@@alin1553 I don't think it has a special name. It is just a theorem arising from the Power of a Point Theorem. It can be proved algebraically (most common) or geometrically.
By synergy property CD=PF, out of interest how do I prove that to be true.
And yes it's sometime since school.
Another commenter pointed out that because of symmetry due to the inscribed rectangle
What inscribed rectangle?, I asked myself.
What was not shown, and should have been was that AB and BE are two sides of an inscribed rectangle. Just dropping in AP then claiming that PF is “obviously” the same length as CD is not adequate. AP needs to also be extended as a cord, perhaps AG, then the cord EG would form the inscribed rectangle, ABEG. Once that is established, then and only then can you stated the symmetry property that defines PF to equal CD.
How do you know that AE passes through the origin-why is that chord different from any other?
@@jeffsmith2208, the centre of the circumscribed circle is the midpoint of the hypotenuse for any right triangle.
Good day. How did you get that AOE are points of the same line? Thanks and regards
We can imagine that the side that has a length of 2 units can be rotated by 90 degrees. We can then form a side length of 6+2=8. This would then form the diameter of the circle , which we could find thru Pythagoras Theorem: (4)^2 + (8)^2 = Root(80)=8.944. But we're not done, radius = 4.472 units -> pi×r^2 = 62.828
Therefore, the Area of the circle is ~62.828
I would set an axis system in point A. Then I'll get 3 points: A(0,0), B(0, 4) and C (6, 6). then I can caluclate radius by using equation of the circle. (x-a)^2 + (y-b)^2 = r^2 where r is radius of circle and a,b are cordinates of the centre of the circle. I have 3 unknow values and 3 points.
Connect A to C ...use proportion Arising from Similarity of triangles...Relationship between chords...Pythagoras
This is much easier. Calculate the side lengths of the triangle ABC and the circumcircle.
Hello,
Would you please explain how you determine that a line from the AB-line center crosses the diameter of the circle perpendicularly; or O, the center of the circle?
There is a theorem which proves that diameter of a circle makes 90 degree
@@Abhishek-hk1on
Hello,
Forget what idiots claim and their theories; everything has theories. I can prove "that circle diameter makes 4.0e-¹°, 2.0e¹°, or 4.800e³°, ETC." People or mathematicians could fool around with mathematics; nevertheless, they could do it with the universe's laws; because it is God's order, and he (God) respects and follows it. God says, "Before the creation, I (God) created logic (physic)," and every matter or manner must follow it. I can also prove that this life is not |absolute| and is an experience of the past-future for the soul of creatures. Studying nuclear cosmology and physic make these phenomena easier to comprehend. Your question or reply doesn't satisfy my question; however, I thank you so much.
Have a beautiful weekend.
@@benjaminkarazi968 Send me the proof then I will believe
@@Abhishek-hk1on
Hello,
Do you feel alright? I do not have to prove anything; nevertheless, the RUclipsr does.
*Much Simpler Solution* - Drop the 2-units line down by 4 units (completing the 6x4 rectangle) and then again by symmetry 2 more units to touch the circle, thereby changing this into a new problem of determining the circum-radius of triangle of sides (a=2+4+2=8), altitude h = 6, b^2=36+36 and c^2 = 36+4. Area = ah/2 = 8x6/2 = 24. abc = (8)(6√2)(2√10) = 192√5. Circum-Radius R = abc / (4 x Area) = 192√5 / (4x24) = 2√5. *Simple*
Sir i had a doubt .
Can we take center on our choice or we have to construct perpendicular from two chords to find the center
Draw the line AC crossing BD at F. Then FBA is similar to FDC, and therefore AB/CD = BF/FD, therefore BF = 4, FD = 2, therefore all the acute angles of these two triangles = pi/4.
Extend BD to meet the circle at E, then ACE is another right-angle triangle (hypotenuse passes through the origin), the angles at C add to pi/2, therefore DCE has acute angles = pi/4. Therefore DE = CD = 2, and BE = 6 + 2 = 8 ==> r = sqrt(20).
Needing some help.
How did we know that AE formed a diameter (intersected the center)?
Thanks
Connect A to C and you have two similar triangles and it is easy to find that they are also Isosceles right triangles. Angle ACD is 45°, so the angle DCE is also 45° and DE is 2 ...
Super. Love watching your videos Sir
Hey pre math :
Your explanation helped to lot of students.
I am also try to help those students ... Who are not affordable good education.
Draw line AC. Notice that this is the diagonal of a 6x6 square, so angle BAC is 45°. This is an angle subtended from chord BC to the circumference, so angle BOC is 45° × 2 = 90°. Therefore 2r^2 = BC^2 = 6^2 + 2^2 = 40, so r = √20 = 2√5.
AB is a chord. So line passing through midpoint of AB will be a diameter. This diameter cuts PD at its midpoint. Since this a circle, by symmetry CD=PF proved...
At around the three minute mark, You say that CD=PF; how do we know this? How do I know to apply symmetry to this diagram? Am I overlooking some simple logic?
Please, and thank You.
B
How it is possible. CD=PF
I don't understand.please explain it.
Symmetry. Inscribe any square or rectangle and any line from it, square to the line of it, to the circle, will be the same on the opposite side of the square/rectangle
(He just didn't complete the rectangle - extend AP to the circle on the right, draw vertical from that point to E and the inscribed rectangle is completed)
I don’t understand either. Drawing vertical line from intersection of AP to the circle may not necessarily go through E. Can u prove it does?
@M Simpson1) I said "inscribed square or rectangle"
Going by picture at 1:57 AB and BE are two lines of an inscribed rectangle. Extend AP to the right so it reaches the circle and you have the 3rd line (parallel to BE). From this new unnamed point, draw a vertical line to E and that's your 4th line, parallel to AB. And there's your complete inscribed rectangle. Symmetry axis is a horizontal line (parallel to BE and AP) through O
@M Simpson Move D to C BD moves up, but also AB to the right and AP down so that P moves to F...
Je ne comprends toujours pas par quelle symétrie CD =PF??????
I solved it in my head, by noticing the angle BAC is 45 degrees. (B is directly up, while C is 6 units up and 6 units right.)
Because the angle BAC is 45 degrees, the arc BC is 90 degrees (double the angle). So BOC is a 45-45-90 right triangle.
The length of BC is sqrt(6*6+2*2) = sqrt(40)
Thus, the radius (OC or OB) is sqrt(40)/sqrt(2) = sqrt(20) = sqrt(4*5) = 2*sqrt(5).
Set circle center as (0,0), since Line AB is 4, so A is (X1, 2), C is (X1+6, 2+2). Put them in two circle equations, get X1 = 4, and r = + sqrt 20.
I solved it in one line in my mind:
Area of triangle ABC is equal to S=12 (side = 4, height = 6).
Another simple formula for area of triangle is S = abc/(4R). Or 12 = abc/(4R) or
R = abc/48, where a, b,c are the sides of triangle.
Sides of the triangle ABC are following: 4, 6sqrt(2) and 2sqrt(10). After multiplication of sides we get R = 2sqrt(5).
So, I have one question. If you only have whole numbers and zero significant digits, how did you get to three significant digits accuracy for your answer?
How can we assume that a chord starting from A, extending through O, will land at point E? Is there some symmetry or rule of circles I’m missing? I mean this diagram isn’t going to be to scale?!
The diameter of a circle subtends a right angle. Conversely, if a chord subtends 90, it is a diameter, goes through the centre.
It's great to remind of the theory.
Regards guys
6-sqrt(r^2-4)=sqrt(r^2-16) solve directly for r=2*sqrt(5) this by projecting to the horizontal and by seeing that the diameter crosses the line size 4
Focus on triangle ABC: Let the 3 sides of ABC be a, b, c, It's easy to get a = BC = 2√10, b=AC=6√2, c=AB=4. From the theorem of cosine: cos B = (a^2 + c^2 - b^2) / 2ac = -1/√10 (since we already know a, b, c). This implies, sin B = 3/√10. Support the R is the radius of the circumscribed circle of triangle ABC. From the theorem of Sine: we have a/sinA = b/sinB = c/sinC = 2R. From 2R = b/sinB, we have R = 2√5. And, the circle in the picture is just the circumscribed circle of triangle ABC. Thus the answer is 2√5.
Alternatively you can see that 2+4 =6 and build back the figure from the circle center :
Take a point O on a grid
from O you can build A by going two step down and four step left
from O you can build B by going two step up and four step left
from O you can build C by going two step right and four steps up
all those path form congruent right angle triangles of sides 2 and 4, hence O is at a distance 2sqrt(5) of A,B and C : we have the radius and the center.
You are 100% right. How is come cd = pf = 2
Analytical solution: define point coordinates A(0, -2), B(0, 2), D(6, 2), C(6, 4). O (circle center) will be at (x, 0) (x axis is perpenticular to AB and has origin at intersection with AB, y axis is AB with y=0 at the middle of AB. O is at the intersection of medians of AB and BC. BC median has equation y=-3x + 12 (BC center is at (3,3)). Solve for y = 0 yields x=4. Then radius is the distance OA (or OB or OC, it does not matter since they're all the same), sqrt(4^2 + 2^2) = 2*sqrt(5).
great presentation!
People don't realize that since A,B,D is a rectangle, it doesn't matter how big it is.
Think about it. Because point A and B touch the circle wall, it's impossible for there to be different gaps above and below it. C,D and P,F will always be the same as long as it's a rectangle and two points touch the wall.
As such, no matter how big or small the rectangle is, A and E will always be the diameter passing through the center.
Another way of solving the problem:
. Overlay the circle with a coordinate plane where the left chord has endpoints (0,0) and (0, -4). The perpendicular bisector of this chord has the formulas y=-2 and runs through the center of the circle. Draw the hypotenuse of the right triangle with sides of length 6 and 2. The coordinates of the endpoints of this chord are (0,0) and (6, 2). Its midpoint is (3, 1), and its slope is 1/3. Its perpendicular bisector is y=-3x+10, and it also runs through the center of the circle. The two perpendicular bisectors intersect at the point (4, -2), which the center of the circle and is a distance of 2*sqrt(5) from (0,0).
How can you prove CD = PF = 2? Just because you made a rectangle in the middle doesnt mean you can say "symmetry" and move the other distance to the bottom at 3:20
How was PF determined to be equal to CD? How was it known that a point halfway between AB makes a line with O that is perpendicular to AB?
Add coordination system,B is(0,0),A is(0,-4),C is(6,2),then becomes more simple.
This proof relies not on Thale's Theorem as stated, but on its converse: the center of the circumcircle of a right triangle lies on its hypotenuse. I think the video would be more helpful if the basic approach of the proof had been outlined up front, thus motivating the steps.
Are we assuming no access to a ruler or tape measure? Because the radius of a circle is half the diameter, so if you know the diameter just divide by 2 and that is the radius or am I just missing something?
Wow this is great... nevertheless I have another interesting solution to this....
......
we know perpendicular bisector of a chord passes through centre 'o' of a circle. Lets coordinate the point B as (0,0) and so C is (6,2)
So perpendicular bisector of AB chord is y = - 2 line..
Slope of BC line = 1/3 => slope of line perpendicular to BC is -3
Midpoint of BC is (6/2,2/2) or (3,1) so equation of the line through midpoint of BC and perpendicular to it is 3x+y-10=0
Intersection of the lines y=-2 and 3x+y-10=0 is the centre which by solving is O(4,-2)
.....note : B is (0,0) ....
Radius= Distance BO = 2✓5
.....thats it
I did not understand the symmetry part in 3:19, how did CD become equal to PF? How did the symmetry come?
@@vashon100 i don’t understand
@@vashon100 not true always. It's only because of the fact it is a right isosceles triangle.
On what basis do you act that the point "O", that is, the center of the circle, is exactly on the line AE?
I approximately almost got the answer just by looking at it though, I am curious of the perfect solution of the problem. I really love this kind of videos. Keep it coming 😊😊😊
I have another solution for mathematicians: For this kind of a problem when you have intersecting chords, and the sides are a, b, c, d just like you named them, you can. ALWAYS find the radius of the circle by this formula : 4r^2 = a^2 + b^2 + c^2 +d^2
Where r is the radius of the circle.
Hint: ^2 means squared e. g
r^2 means r squared.
You can thank me later!
Nice!
What symmetry property. Is invoked to show PF=2 ?
It's as simple using the coordonates of the point, the origin being B: B(0,0), A(0, -4), C(6,2), and using the circle equation (x-a)^2 + (y-b)^2= r.
R being the radius and a,b the coordinates of the center.
1) B being on the circle: a^2 + b^2= r^2
2) A being on the circle: a^2 + (b+4)^2= r^2. Wich gives by substracting 1) b=-2 and a^2= r^2 - 4
3) C being on the circle: (6-a)^2 + (2-b)^2= r^2 ==> (6-a)^2 + 16= a^2+4 from 2). So (6-a)^2= a^2-12, i.e. 36 + a^2 - 12a= a^2 -12 ==> a=4
4) r^2= 16+4=20 so r= sqrt(20)= 2*sqrt(5).
Much simpler solution... if you draw AC, it works out that you have two similar triangles, and that angle BAC is 45 degrees. BC is therefore the chord of a 90 degree radial angle, length 2 root 10. That leads to a radius of 2 root 5.
Consider the points A,B,C,are points on the circle. Let O be the centre say (x,y).
Co ordinates of A(0,0)B(0,4) and C(6,6)
Solve for OA=OB,we get y=2
Solve for OB=OC we get x=4
Hence radius= ✓4^2+2^2=✓20=2✓5 is correct
this is the best solution i think
O lower than B at 2.
Middle of BC higher at 1 and righter at 3 fron B.
O lower than that middle at 3.
Right shift of O 2/6 of that high and equal 1.
Takes O lower at 2 and righter at 4.
So radius is √(2^2+(3+1)^2)=√20=2√5
Seems thais way a bit shorter...
Sorry if my english poor - learn it at other century :) and almost don't use in life.
Awesome channel.... I love those math puzzles