It's even better when you try to come up with different ways to get the result. I used the law of cosines and the center angle/circumference angle theorem, as well as some distance formula.
@@trancozin130 You only need to consider the circle and the right square. Because it's touching the circle on both of the right corners, the rightmost extremity of the circle has to be right on the horizontal symmetry axis of the right square. If you imagine drawing a horizontal line through the middle of the square, you'll see that it goes through the right extremity of the circle. There's probably some proof for this, but to me it makes intuitive sense so it's hard to explain...
@@HR3EEE the square is made of 4 straight, perpendicular, congruent lines. this means that the square is always "centered" if two of its (non opposite) corners are touching the edge of the circle. so you know the length of the square's side is 4, and thus half of that is 2. (were the opposite corners touching, you would instead have k be half the diagonal.)
Andy, It would be so cool to see how you develop a curriculum for all the "-not-so-often-used" formulae. I sit here befuddled by the problem and very often your method of attack seems new to me. Keep up the great work!
This was really cool to do first and then watch you do it. I took a more geometric approach that gave me similar equations to solve. Ignoring all of the stuff I drew that was unhelpful, here’s what worked for me: All of the squares have the same area of 16. That means their side lengths are sqrt(16)=4. Three points define a unique circle. The center must be equidistant from all three points. That distance is the radius. I drew an approximate center with radii out to each point. Since the two points on the right are exactly vertical to each other, this center would be vertically equidistant between the two, meaning it was vertically 4/2=2 units higher than the lower right point. I also extended the “grid” formed by the squares, specifically horizontal line segments from the bottom-left square over to the right and across the bottoms of the two vertically middle squares. Then, I dropped a vertical line from the center down to the lowest horizontal line segment I drew. This formed two right-angle triangles, both with radii as the hypotenuses. The triangle on the left had a height of 6 and an unknown width of x. The triangle on the right had a height of 2 and an unknown width of 16-x. Using the Pythagorean Theorem, I was able to set r^2 equal to two different statements containing x. r^2=(6^2)+(x^2) r^2=(2^2)+((16-x)^2) I then set those equal to each other and solved for x. (6^2)+(x^2)=(2^2)+((16-x)^2) 36+(x^2)=4+256-32x+(x^2) -224=-32x x=7 I plugged x back into one of the equations to find r^2. r^2=(6^2)+(7^2) r^2=36+49 r^2=85 And then the final answer! A=85(pi) square units
If you start with a little geometry and draw the perpendicular bisector of two of the circle's chords, you find the center of the circle very easily and from there a little use of Pythagorean theorem gives you the radius squared. Multiply by Pi and voilà :)
Did you actually work it out like that? I totally concede that I might miss an obvious solution using only the Pythagorean theorem here, but I think you need more tools. I also thought it should be quite easily solvable and wanted to comment the easier solution to the problem here, but in the process of working it out I found out I needed some more tools and I have an argument why Pythagoras isn't enough. Assume P1,P2 and P3 are given as coordinates and lie on a circle. We can now do the perpendicular biscetion you talked about of the lines P1 P2 and P2 P3. Let's call the point the perpendicular lines cross the chord of the Circles S1 and S2. Now we have right triangles, where the radius is one length, the distance from the middle point to S1 and S2 and the Distance between S1 or S2 to P1, P2 or P3. The system of equations we get from Pythagoras now is not uniquely solvable. It cannot be, since we have ONLY used the distances between P1,P2 and P2,P3. These distances are not enough to uniquely define a circle. We need the information about the angle. If I use the law of cosines and compute the angle where the lines meet at P2, we can actually work it out, since we can quite easily get the angles in the right triangles between S1 P2 and the middle point. If we have these angles, we can get the radius, since we know two angles and the Distance between P2 and S1. But honestly the calculation is a pain and definitely harder than what was shown in the video. If I'm wrong, I would be super interested in a solution that works the way you have described it, but I found it to be way more tricky than it appears at first glance.
@@maxwellsdemon10 I think your issue is only one of method. You are generalising the problem to find a solution for all possible P1, P2 and P3 and then try to plug in the specific P1, P2, P3 of the problem to find the solution of the problem. So basically you are building a nuclear missile to kill a mosquito :D If you work step by step using the specifics of the problem, it is dramatically easier. Let's have the bottom left contact point with the circle be P1 et let's set it at (0;0), making P2 (16;4) and P3 (16;8) since squares of area 16 have a side of length 4 (and I am using horizontal and vertical for my x and y axis). From this you can calculate the middles points S1 (8;2) and S2 (16;6). Very obviously the perpendicular bisector of the chord P2-P3 will be horizontal since the chord is vertical, and we know it will go through S2 and the center of the circle (let's call it C )so we know both have the same Y coordinate : 6. The other perpendicular bisector is slightly more complicated as it is slanted, but we know it's slant will be the negative inverse of the one of the chord P1-P2 since it's perpendicular. The slant of of P1-P2 is 4/16 or 1/4, so the slant of or perpendicular bisector is -4. Things work out very nicely for us because the bisector goes through S1(8;2) and we want to go up to the center C (?;6) and that's just 4 above, so we just need to go 1 back on the x axis, so C is at (7;6). Note that these previous two verbose paragraphs with coordinates take 10 seconds on graph paper if you draw the problem. So now with C, S2 and P3 we can go pythagoring: C-S2 is length 9, S2-P3 length 2, 9 squared plus 2 squared is 85 witch is the radius squared since C-P3 is a radius of the circle. so 85 Pi is our answer for the area of the circle! How exciting I guess :D
Oh how the universe aligns! Idk if you will believe me, but I am a drywaller. On monday I have to make a decorative circle in a ceiling and shit needs to hit few specific spots. This video is exactly what I will be using. Thanks yt recommendations!
Very cool solution! Also make the algebra a little less spicy by reading k=2 from the coordinate plane at 1:14 as that's the only way a square can fit like that
I was thinking the same thing. And then you only need two points, which means, there is only one way a single square can fit like that. You can also see that we don't need 3 points, because once you know k = 2, (0, 0) and (0, 4) give the exact same information.
The rightmost square is centred symmetrically on the horizontal diameter of the circle (this has to be so since the chord it makes with the circle is vertical, but can be made clearer by drawing additional mirror flips of the other squares.) Given this symmetry, k is obviously 2 by inspection.
GENIUS I’m over here thinking how we can use arcs and angles to try and figure out the area. It never even occurred to me to plot the circle on a graph
That was a spicy application of co-ords to solve this problem. I was realizing you need to solve for r (or r^2) but couldn't figure out how to start. Co-ords to the rescue.
is there an alternate way of doing this like using the touching points of the squares to form a triangle and then using lengths of the triangle to determine the size of the circle.
I got there by adding a square in the top left of the circle and making a fourth point. After that I could make multiple lines and calculate the radius that way. But your solution is very nice too, even nicer I'd say.
I used the law of sines to solve it! View the three points touching the circle as vertices to a triangle. Then the circle is the circumcircle of the triangle. Let A be the top right vertex, B be bottom right, and C be bottom left. By law of sines, a/sin(A)=2R where R is the radius of the circumcircle and a=length of BC. sin(A) is easy to calculate: imagine a right triangle by extending AB downwards by another 4 units to point D. Then ADC is a right triangle, and we get AD=8, DC=16, so by Pythagorean AC=8sqrt(5), so sin(A)=opp/hyp=DC/AC=2/sqrt(5). Now we need a=BC, but this can be solved by applying Pythagorean on triangle BDC, since BD=4, DC=16, we have a=BC=4sqrt(17). Combining all these information together, we have R=a/2sin(A)=4sqrt(17)/(4/sqrt(5))=sqrt(85). The area of the circle is this πR^2=85π. Using this method I was able to mental math it in a few minutes 🤓
Bless your efforts, Professor. I follow you from Syria. I really enjoyed watching your videos. You explain to us the solution in a very wonderful way Professor, I want to ask you a request. I just want you to add to your videos the feature of translation into Arabic, because I am not fluent in the English language. All I have is Arabic. I even used Google Translate to express what I want to you. All my love to you, Professor ❤️
Whoa, I did this a completely different way by using the inscribed angle theorem to compute the arc angle between the two points on the right, then using that to compute the radius of the circle. I think you got a cleaner form than me: (4pi)/(sin^2(arctan(1/2)-arctan(1/4)) isn't quite as clean as 85pi, but they're equal!
I took the surveyor approach (being a surveyor and surveying instructor). Essentially took the bottom side of the bottom left square, extended it to right to intersect right side of rightmost square extended down. Established perpendicular bisector to left of rightmost square. From this, established right trapezoid with bottom base length of 16, right side perpendicular to base with length of 6, side from right side bisector left to radius point with a length sqrt(R^2-4), and a diagonal left side with distance of R. From radius point, construct line perpendicular to bottom base. This leads to a right triangle with hypotenuse of R, one side of 6, and another side of 16-sqrt(R^2-4). Plug into Pythagorean Theorem and end up with R^2=85. Area=85π.
You can also figure out the y-coordinate of the circle centre by symmetry, since the rightmost square is touching the circle at two points. I'm not sure how you would prove it, but that must mean the horizontal line bisecting the circle must be exactly equidistant to each of the two points, and since the side length is 4, the centre y-coord must be 2.
I never thought to use coordinates and the equation of a circle! I was able to solve it by using the chords of a circle theorem. I knew we had a chord with length 4 on the right. If we extend the vertical line on the left, we get another chord with length 12. The diameter is a perpendicular bisector of both chords. From there I used some Pythagorean theorem and got the same answer. I love that some of these problems can be solved in multiple ways. Thanks for sharing!
I think it's not so obvious to tell that the second chord has a lenght of 12 Instead you can draw a second chord from bottom left to top right point. The middle of the chord will be in the bottom left corner of a middle square and the perpendicular to that chord will intersect top side of that square in the middle. From that point it's quite simple to calculate radius using Pythagorean theorem.
I did this defining the bottom left point on the bottom left square as the origin, finding the equations of the lines from that point to the other 2 points on the circumference, then finding the equations of 2 lines perpendicular to those 2 lines that pass through the midpoint (basically the perpendicular bisectors), then by finding the intersection point (7,6) i can calculate the radius by calculating the distance to the origin, since the origin is on the circumference of the circle (pythagoras) and then use that to calculate the area. Very exciting
You can also solve it by using the 3 points as vertices of a triangle. The circle is then the circumcircle of the triangle. The perpendicular bisectors come out nicely once we assign coordinates to these points. The slopes of these perpendicular bisectors are quite nice, and their intersection (the center of the circle) is an integer point just like the the vertices of the triangle. So then the square of the radius can be obtained using Pythagorean theorem. I used the top-right vertex and found 85 = 9^2 + 2^2.
Well.. I way overcomplicated this problem. I drew an inscribed triangle, used the distance formula to find each side length, used law of cosines to find one of the angles in the triangle, used law of sines to find the radius, and finally calculated the area. I got the same answer but I did wayyyy more work lol
This one took me a while. I took a totally different approach, which was very simple in setup, but took quite a bit of working out. I used trig. It's kind of hard to explain, but basically, using the coordinate system you use (though the way I did it, I didn't need a coordinate system), I'm looking for x, the distance from (0, 2) to the center. I can also find alpha, the angle between (0, 2) to (0, -14) and (0, 2) to (4, -14). I also note the unknown angle theta, the angle between the center to (0, 2) and the center to (0, 0). From these relationships, it's not too hard to see tan(alpha) = 1/4 and tan(theta) = 2/x. The last piece of the pie is noticing (through some angle relationships) that the angle between the center to (0, -14) and the center to (4, -14) is 2alpha + theta. This gives tan(2alpha + theta) = 6/(16 - x). I then used the tan sum and tan double angle identities to get the last equation in terms of tan(alpha), tan(theta), and x. Substituting, I can get an equation totally in terms of x. After a lot of algebra, I got 0 = 8x^2 - 8x - 576. Using our old friend the quadratic formula, there are two solutions: -8 and 9. It's obviously not -8, since it's a length, so it's 9! From that, I could get the radius with Pythagorus: 9^2 + 2^2 = r^2.
Nice solution with algebraic approach and Cartesian axies. I think I found faster solution:) Solving area of triangle (Pitagoras for missing two). Then I am using alternative area solver for triangle A=abc/4R which is direct way to get R, what do ou thing about it?
Question for you. At around 2:40, you have k^2 = (4-k)^2 Can you take the square root of both sides in this situation? The results still comes out to K = 2 but I seem to remember that there are some situations where you can take the square root of both sides of an equation, but there are other situations where you cannot. Is that right?
I’d like to share another geometric approach to this question. By connecting the three red dots you get a triangle which the circle is its circumcircle And simply by calculating the area and sides fo the triangle gives you the radius of the circumcircle Easier to formulate and calculate AT = area of triangle R = radius of the circle (a,b,c) = three sides of the triangle (A,B,C)=the three angles of the triangle for 0.5*ab*sinC = AT & 2*R*sinC = c We have (abc)/(4*AT) = R = sqrt(85) hence area of circle is 85*pi
i completely forgot about coordinate circle solving! very clever!
Me too!
As did I. I immediately went into forming triangles. But lo and behold, coordinates to the rescue! How. Exciting.
Me too!^
I legit had that as my first thought, lol.
Same here, great to refresh the mind.
The double box was very exciting :)
The rare and elusive double box appears.
Underful hbox badness 10000
this was one of the best ones so far. I absolutely loved the idea of putting it on an axis - first time seeing a solution like this. how exciting
how exciting
hpw exciting
how exciting
how exciting
how exciting
how exciting
Your channel made me like math and look at it differently. More like fun puzzles.
same
Using coordinated plane here is really clever
really fun doing honestly
Mind Blown. It's been so long but you still make it look so easy. This is one of my favorites. Rock On
These videos make me feel speechless, he is not showing how to use numbers and formulas, he is teaching to think and to solve problems.
HOW EXCITING!
It's even better when you try to come up with different ways to get the result. I used the law of cosines and the center angle/circumference angle theorem, as well as some distance formula.
I love how the operations are in a different color. Visually makes things easier to follow, thanks!
I love the way that you work the equations on the screen. Very smooth and easy to follow. How exciting!
How spicy! 🌶️
Loved this one, Andy. Ingenious solution!
How spicy.
Because of the right square touching the circle twice, you can already deduce that k=2
Can you explain?
@@trancozin130 You only need to consider the circle and the right square.
Because it's touching the circle on both of the right corners, the rightmost extremity of the circle has to be right on the horizontal symmetry axis of the right square. If you imagine drawing a horizontal line through the middle of the square, you'll see that it goes through the right extremity of the circle. There's probably some proof for this, but to me it makes intuitive sense so it's hard to explain...
@@paulvansommerenwe’re gonna need some back up explaining here lol.
@@HR3EEE the square is made of 4 straight, perpendicular, congruent lines. this means that the square is always "centered" if two of its (non opposite) corners are touching the edge of the circle. so you know the length of the square's side is 4, and thus half of that is 2. (were the opposite corners touching, you would instead have k be half the diagonal.)
@@pikminman13I dont understand, you mean after the circle is put on the plane?
You never fail to impress me. I am envious.
Very good. Enjoyed the use of the circle formula.
Food is always better when spicy, at least a little.
Food is always worse when spicy, no exceptions.
@@ExzaktVid Unlikely you will get a heart.
what about candy? Do we count that to food?
@@qyxyp Sure. Why not?
@@richardl6751 because it‘s not better when it‘s spicy mostly
I would’ve never thought about using the coordinates for this problem. You’re brilliant!
Please keep up making those short and intuitive problems.
The channel i relax to everytime he uploads. Thank you andie
This might be one of my favorite videos of yours. How Exciting!
New subscriber here, very spicy work! Im a CAD//CAM guy and circles are my life. I always enjoy a good circle riddle.
I dont know why, but this one was very satisfying to watch 🤩
That's a spicy assignment.
Andy, It would be so cool to see how you develop a curriculum for all the "-not-so-often-used" formulae. I sit here befuddled by the problem and very often your method of attack seems new to me. Keep up the great work!
This was really cool to do first and then watch you do it. I took a more geometric approach that gave me similar equations to solve.
Ignoring all of the stuff I drew that was unhelpful, here’s what worked for me:
All of the squares have the same area of 16. That means their side lengths are sqrt(16)=4. Three points define a unique circle. The center must be equidistant from all three points. That distance is the radius. I drew an approximate center with radii out to each point. Since the two points on the right are exactly vertical to each other, this center would be vertically equidistant between the two, meaning it was vertically 4/2=2 units higher than the lower right point. I also extended the “grid” formed by the squares, specifically horizontal line segments from the bottom-left square over to the right and across the bottoms of the two vertically middle squares. Then, I dropped a vertical line from the center down to the lowest horizontal line segment I drew. This formed two right-angle triangles, both with radii as the hypotenuses. The triangle on the left had a height of 6 and an unknown width of x. The triangle on the right had a height of 2 and an unknown width of 16-x. Using the Pythagorean Theorem, I was able to set r^2 equal to two different statements containing x.
r^2=(6^2)+(x^2)
r^2=(2^2)+((16-x)^2)
I then set those equal to each other and solved for x.
(6^2)+(x^2)=(2^2)+((16-x)^2)
36+(x^2)=4+256-32x+(x^2)
-224=-32x
x=7
I plugged x back into one of the equations to find r^2.
r^2=(6^2)+(7^2)
r^2=36+49
r^2=85
And then the final answer!
A=85(pi) square units
This is going to help me so much with my algebra EOC on Thursday, Thanks! That was spicy!
If you start with a little geometry and draw the perpendicular bisector of two of the circle's chords, you find the center of the circle very easily and from there a little use of Pythagorean theorem gives you the radius squared. Multiply by Pi and voilà :)
Did you actually work it out like that?
I totally concede that I might miss an obvious solution using only the Pythagorean theorem here, but I think you need more tools.
I also thought it should be quite easily solvable and wanted to comment the easier solution to the problem here, but in the process of working it out I found out I needed some more tools and I have an argument why Pythagoras isn't enough.
Assume P1,P2 and P3 are given as coordinates and lie on a circle.
We can now do the perpendicular biscetion you talked about of the lines P1 P2 and P2 P3. Let's call the point the perpendicular lines cross the chord of the Circles S1 and S2.
Now we have right triangles, where the radius is one length, the distance from the middle point to S1 and S2 and the Distance between S1 or S2 to P1, P2 or P3.
The system of equations we get from Pythagoras now is not uniquely solvable.
It cannot be, since we have ONLY used the distances between P1,P2 and P2,P3.
These distances are not enough to uniquely define a circle. We need the information about the angle.
If I use the law of cosines and compute the angle where the lines meet at P2, we can actually work it out, since we can quite easily get the angles in the right triangles between S1 P2 and the middle point. If we have these angles, we can get the radius, since we know two angles and the Distance between P2 and S1.
But honestly the calculation is a pain and definitely harder than what was shown in the video.
If I'm wrong, I would be super interested in a solution that works the way you have described it, but I found it to be way more tricky than it appears at first glance.
@@maxwellsdemon10 I think your issue is only one of method. You are generalising the problem to find a solution for all possible P1, P2 and P3 and then try to plug in the specific P1, P2, P3 of the problem to find the solution of the problem. So basically you are building a nuclear missile to kill a mosquito :D If you work step by step using the specifics of the problem, it is dramatically easier.
Let's have the bottom left contact point with the circle be P1 et let's set it at (0;0), making P2 (16;4) and P3 (16;8) since squares of area 16 have a side of length 4 (and I am using horizontal and vertical for my x and y axis). From this you can calculate the middles points S1 (8;2) and S2 (16;6). Very obviously the perpendicular bisector of the chord P2-P3 will be horizontal since the chord is vertical, and we know it will go through S2 and the center of the circle (let's call it C )so we know both have the same Y coordinate : 6.
The other perpendicular bisector is slightly more complicated as it is slanted, but we know it's slant will be the negative inverse of the one of the chord P1-P2 since it's perpendicular. The slant of of P1-P2 is 4/16 or 1/4, so the slant of or perpendicular bisector is -4. Things work out very nicely for us because the bisector goes through S1(8;2) and we want to go up to the center C (?;6) and that's just 4 above, so we just need to go 1 back on the x axis, so C is at (7;6).
Note that these previous two verbose paragraphs with coordinates take 10 seconds on graph paper if you draw the problem.
So now with C, S2 and P3 we can go pythagoring: C-S2 is length 9, S2-P3 length 2, 9 squared plus 2 squared is 85 witch is the radius squared since C-P3 is a radius of the circle. so 85 Pi is our answer for the area of the circle! How exciting I guess :D
well what if it’s not to scale
It's amazing!
I'm a Math teacher in Brazil and I really love your videos
Soo cool ! I almost forgot how fun can it be.... Thanks for the great video!
The solution brought a smile on my face! Not exactly the solution, but the approach!
Fantastic and making it look/sound easy
Oh how the universe aligns!
Idk if you will believe me, but I am a drywaller. On monday I have to make a decorative circle in a ceiling and shit needs to hit few specific spots. This video is exactly what I will be using. Thanks yt recommendations!
This is the first time I saw you put a box around a box.
How exciting.
Clear step by step instructions with visualisations.
the double box was really spicy; loved it
Love it! Thanks for a brilliant solution, although the equation of a circle is new to me. Live and learn.
Really fun watch, pretty spicy solution you displayed
I didnt think of this, very nice solution!
Very cool solution! Also make the algebra a little less spicy by reading k=2 from the coordinate plane at 1:14 as that's the only way a square can fit like that
I was thinking the same thing. And then you only need two points, which means, there is only one way a single square can fit like that. You can also see that we don't need 3 points, because once you know k = 2, (0, 0) and (0, 4) give the exact same information.
The rightmost square is centred symmetrically on the horizontal diameter of the circle (this has to be so since the chord it makes with the circle is vertical, but can be made clearer by drawing additional mirror flips of the other squares.)
Given this symmetry, k is obviously 2 by inspection.
Nice. Afaik there is a formula for this and I expected you to use it, but solving it using the circle equation was way more intuitive.
"How Exciting" gets me everytime 😭😂
So genius ❤ love it!
GENIUS
I’m over here thinking how we can use arcs and angles to try and figure out the area. It never even occurred to me to plot the circle on a graph
nice explnation for equations of a circle, can you upload a full video explainin the equation?
That was a spicy application of co-ords to solve this problem. I was realizing you need to solve for r (or r^2) but couldn't figure out how to start. Co-ords to the rescue.
is there an alternate way of doing this like using the touching points of the squares to form a triangle and then using lengths of the triangle to determine the size of the circle.
I feel like go back to school!!! haha nice one Andy.
never seen it solved like this, awesome!
I love your videos! ♥
"How. Exciting." Gets me every time
Awesome content, thanks!!
I've been out of school for 8 years and I'm somehow interested in watching this math video
I got there by adding a square in the top left of the circle and making a fourth point. After that I could make multiple lines and calculate the radius that way. But your solution is very nice too, even nicer I'd say.
I used the law of sines to solve it! View the three points touching the circle as vertices to a triangle. Then the circle is the circumcircle of the triangle. Let A be the top right vertex, B be bottom right, and C be bottom left. By law of sines, a/sin(A)=2R where R is the radius of the circumcircle and a=length of BC. sin(A) is easy to calculate: imagine a right triangle by extending AB downwards by another 4 units to point D. Then ADC is a right triangle, and we get AD=8, DC=16, so by Pythagorean AC=8sqrt(5), so sin(A)=opp/hyp=DC/AC=2/sqrt(5). Now we need a=BC, but this can be solved by applying Pythagorean on triangle BDC, since BD=4, DC=16, we have a=BC=4sqrt(17). Combining all these information together, we have R=a/2sin(A)=4sqrt(17)/(4/sqrt(5))=sqrt(85). The area of the circle is this πR^2=85π.
Using this method I was able to mental math it in a few minutes 🤓
Bless your efforts, Professor. I follow you from Syria. I really enjoyed watching your videos. You explain to us the solution in a very wonderful way Professor, I want to ask you a request. I just want you to add to your videos the feature of translation into Arabic, because I am not fluent in the English language. All I have is Arabic. I even used Google Translate to express what I want to you. All my love to you, Professor ❤️
I would have thought to find the center and then the radius from that but this is cool
How exciting bro.. Really amazing
Whoa, I did this a completely different way by using the inscribed angle theorem to compute the arc angle between the two points on the right, then using that to compute the radius of the circle. I think you got a cleaner form than me:
(4pi)/(sin^2(arctan(1/2)-arctan(1/4)) isn't quite as clean as 85pi, but they're equal!
Nice! Very clear, thank you.
Damn! Learned something new today. 👍👍
I took the surveyor approach (being a surveyor and surveying instructor). Essentially took the bottom side of the bottom left square, extended it to right to intersect right side of rightmost square extended down. Established perpendicular bisector to left of rightmost square. From this, established right trapezoid with bottom base length of 16, right side perpendicular to base with length of 6, side from right side bisector left to radius point with a length sqrt(R^2-4), and a diagonal left side with distance of R. From radius point, construct line perpendicular to bottom base. This leads to a right triangle with hypotenuse of R, one side of 6, and another side of 16-sqrt(R^2-4). Plug into Pythagorean Theorem and end up with R^2=85. Area=85π.
You can also figure out the y-coordinate of the circle centre by symmetry, since the rightmost square is touching the circle at two points. I'm not sure how you would prove it, but that must mean the horizontal line bisecting the circle must be exactly equidistant to each of the two points, and since the side length is 4, the centre y-coord must be 2.
Wow, super clever. So Good!
I've never heard about the circle equation
Very interesting, I learned something
yo I appreciate the videos! Also, I've always wanted to know what program you use to visualize with.
I never thought to use coordinates and the equation of a circle!
I was able to solve it by using the chords of a circle theorem. I knew we had a chord with length 4 on the right. If we extend the vertical line on the left, we get another chord with length 12. The diameter is a perpendicular bisector of both chords. From there I used some Pythagorean theorem and got the same answer.
I love that some of these problems can be solved in multiple ways. Thanks for sharing!
I think it's not so obvious to tell that the second chord has a lenght of 12
Instead you can draw a second chord from bottom left to top right point. The middle of the chord will be in the bottom left corner of a middle square and the perpendicular to that chord will intersect top side of that square in the middle. From that point it's quite simple to calculate radius using Pythagorean theorem.
This was so spicy. It brought memories of High School where I took IB Maths.
He makes this look so simple.
i've learned this subject before my classmates did. how exciting.
That gives good vibes dude😅❤
You Can Also make a triangle between the points touching the circle and use trigonometry to solve it
@AndyMath - How do you create these spicy solutions in the form of Animation? What software are you using? Would anyone else be able to help me?
I love this stuff
I did this defining the bottom left point on the bottom left square as the origin, finding the equations of the lines from that point to the other 2 points on the circumference, then finding the equations of 2 lines perpendicular to those 2 lines that pass through the midpoint (basically the perpendicular bisectors), then by finding the intersection point (7,6) i can calculate the radius by calculating the distance to the origin, since the origin is on the circumference of the circle (pythagoras) and then use that to calculate the area. Very exciting
two boxes around it?? now THAT is spicy
This might be the most exciting one yet.
Wow ² + amazing=(How, Exciting)
I wish I had this channel when I was in college...
That was pretty satisfying 😁
You can also solve it by using the 3 points as vertices of a triangle. The circle is then the circumcircle of the triangle. The perpendicular bisectors come out nicely once we assign coordinates to these points. The slopes of these perpendicular bisectors are quite nice, and their intersection (the center of the circle) is an integer point just like the the vertices of the triangle. So then the square of the radius can be obtained using Pythagorean theorem. I used the top-right vertex and found 85 = 9^2 + 2^2.
That's the way I did it. A bit easier I think (at least in this case where the numbers are nice and neat).
The exciting sequel to "2 girls 1 cup"
Hello Andy, i was wondering if there was a way to send you questions. If theres one, can you link it in the replies section of this comment? Thanks.
The moment you really start learning math is when you start having fun doing math
had me saying wow at the end of it
Well.. I way overcomplicated this problem. I drew an inscribed triangle, used the distance formula to find each side length, used law of cosines to find one of the angles in the triangle, used law of sines to find the radius, and finally calculated the area. I got the same answer but I did wayyyy more work lol
This one took me a while. I took a totally different approach, which was very simple in setup, but took quite a bit of working out. I used trig. It's kind of hard to explain, but basically, using the coordinate system you use (though the way I did it, I didn't need a coordinate system), I'm looking for x, the distance from (0, 2) to the center. I can also find alpha, the angle between (0, 2) to (0, -14) and (0, 2) to (4, -14). I also note the unknown angle theta, the angle between the center to (0, 2) and the center to (0, 0). From these relationships, it's not too hard to see tan(alpha) = 1/4 and tan(theta) = 2/x. The last piece of the pie is noticing (through some angle relationships) that the angle between the center to (0, -14) and the center to (4, -14) is 2alpha + theta. This gives tan(2alpha + theta) = 6/(16 - x). I then used the tan sum and tan double angle identities to get the last equation in terms of tan(alpha), tan(theta), and x. Substituting, I can get an equation totally in terms of x. After a lot of algebra, I got 0 = 8x^2 - 8x - 576. Using our old friend the quadratic formula, there are two solutions: -8 and 9. It's obviously not -8, since it's a length, so it's 9! From that, I could get the radius with Pythagorus: 9^2 + 2^2 = r^2.
Imagine doing this in a test and forgetting to multiply it by pi in the end
That was an spicy approach XD
Ok, this one is really cool
Nice solution with algebraic approach and Cartesian axies.
I think I found faster solution:)
Solving area of triangle (Pitagoras for missing two). Then I am using alternative area solver for triangle A=abc/4R which is direct way to get R, what do ou thing about it?
Coordinates for circles? Holy.
I love this channel
These are both amazing and depressing, mostly because I realize how little I really know
Love a bit of spicy math ☺
Question for you. At around 2:40, you have k^2 = (4-k)^2
Can you take the square root of both sides in this situation? The results still comes out to K = 2 but I seem to remember that there are some situations where you can take the square root of both sides of an equation, but there are other situations where you cannot. Is that right?
I’d like to share another geometric approach to this question.
By connecting the three red dots you get a triangle which the circle is its circumcircle
And simply by calculating the area and sides fo the triangle gives you the radius of the circumcircle
Easier to formulate and calculate
AT = area of triangle
R = radius of the circle
(a,b,c) = three sides of the triangle
(A,B,C)=the three angles of the triangle
for 0.5*ab*sinC = AT & 2*R*sinC = c
We have (abc)/(4*AT) = R = sqrt(85)
hence area of circle is 85*pi
Can you do maths past papers or something or to help us students 😅
tryly awesome and simple
Where do you find these challenges?