Yeah, if you don't verify mathematically, you can't really know that what visually "looks right" is right. A good example of this is the disappearing chocolate piece when you slice a chocolate bar the right way.
This is because we saw how the biggest triangle(big+small) has alpha and theta angle along with a right angle. And we know the sum of angles in a triamgle is 180. So we get Alpha + theta + 90(right angle) =180 Since we know both triangles have a right angle and theta angle, we can concludw that both triangles contain the alpha angle. Hoped this made sense
@@mrmailman5561not necessarily the same he’s using theta to show those particular angles and the rest is all mix of “sohcahtoa” and other ratios between those angles and their relationship to the circle they are a part of. Edit: aaaaaaaaaaaaand I just saw you got the answer already lol sry I’m hi
I'm so happy I got part of the solution correct just by eye-balling it, I watched enough of your videos to instinctively connect the 2 lines of the circles from the center and I also saw the small right triangle with the Y section, your videos really are helpful in helping with problem solving, Andy, thanks a lot
well yeah... that's a nice way to verify the approximate answer but not more i remember seeing the similar video where it was quite obvious to me that the answer was 3,5 based on a grid. but it turned out to be something like 1,23π you really never know...
Quick solution using homothety: Let the red and yellow semicircles be tangent at T and touch the black semicircle at A and B, respectively. Let |AT| = x. Consider the homothety from T (with factor -1.5) sending A to B and notice it maps the red semicircle to the yellow one => A, B and T are on a line and |TB| = 1.5x, so |AB| = x + 1.5x = 2.5x. Now consider the homothety from A sending T to B and notice it maps the red semicircle to the black semicircle, meaning the ratio of their diameters is the ratio of chords AT and AB, namely 2.5x / x = 2.5 hence the unknown radius is 2.5 * 2 = 5.
It would be interesting if you showed if this diameter being the sum of both diameters is a coincidence here or true for any combination of two diameters.
I did the math using arbitrary diameters and all complicated math gets cancelled out until there is only the sum of the diameters! Pretty much the same way he did it, just using a,b instead of 2,3.
It is indeed true for any combination of diameters. Assume the two smaller semi-circles have radii a and b (diameters 2a and 2b). We can show that larger semi-circle has diameter 2a + 2b. We'll divide the diameter into segments of length a, c, b, and y (see diagram at 2:45. Since c is leg of right triangle, whose other leg = b, and hypotenuse = a+b, we get c² + b² = (a+b)² c² = a² + 2ab Now we calculate y using similar triangles, as shown in video: y/b = b/(a+c+b) y = b²/(a+b+c) Now we calculate diameter of larger semi-circle: d = a + c + b + y d = a + b + c + b²/(a+b+c) d = ((a+b+c)² + b²) / (a+b+c) d = ((a² + b² + c² + 2ab + 2ac + 2bc) + b²) / (a+b+c) Replace c² with a² + 2ab d = (a² + b² + (a² + 2ab) + 2ab + 2ac + 2bc + b²) / (a+b+c) d = (2a² + 2b² + 4ab + 2ac + 2bc) / (a+b+c) d = (2 (a²+b²+2ab) + 2c(a+b)) / (a+b+c) d = (2 (a+b)² + 2c(a+b)) / (a+b+c) d = 2 (a+b) (a+b+c) / (a+b+c) d = 2 (a+b) d = 2a + 2b = (sum of diameters of smaller semi-circles)
Awesome solution with similar triangles! How I did it: Equation of the circle is (x-h)² + y² = r². Circle goes through (0,0) so plug in x=0 and y=0 to get h = r. Thus the equation is (x-r)² + y² = r². Using the same reasoning as you did to find the 3-4-5 triangle, the coordinates of the bottom of the yellow semicircle are (3,0), and thus the rightmost point is (4.5,1.5). Plug in x=4.5, y=1.5 into the equation: (4.5-r)² + 1.5² = r². This becomes an easy linear equation, and r=2.5. Thus the diameter d = 2(2.5) = 5. This way feels more down and dirty compared to your elegant solution!
I love how you always knew it was 5 but you really took the time to teach us the geometry rules that prove its 5 and how we know that. its really finding complex structures within simple questions that does it for me thanks andy
Your initial intuition inspired another solution: Suppose you extend both the yellow and orange semi-circles to be complete circles. Now imagine the yellow circle orbiting clockwise around the orange circle, keeping tangent. During that orbit the distance between their centers obviously remains constant, namely 2.5 units (sum of the radii). The sum of their diameters is just double that, namely 5 units. When the yellow circle's center is on the bottom line of the large semicircle, it's still 5 units, so 5 must be the diameter of the large semicircle.
Interesting thinking but this is not a full solution. After the yellow circle is rotated around the smaller one so its center lies on the bottom line, how do you know its right edge lines up perfectly with the corner of the larger semicircle?
I ended up using “perpendicular bisector to any chord passes through the center of the circle” and drew bisector on the hypotenuse of 4.5 and 1.5. Your method is simpler though.
There's another (2,1.5,2.5) rightwing triangle between the center of the black circle, the rightmost point where yelllow circle touches the black circle and the point on the line that is distance y from the right. From this other triangle together with the first one, we can derive the diameter of the large circle.
I have to admit that the prob is really fun. I did it using the trigonometric function and Pythagoras theorem. My solution is similar to the video, except the last step. After found the 4.5 side, I use Pythagoras theorem to find the hypotenuse^2 of the big triangle "4.5^2+1.5^2=22.5". Followed by the trigonometric function in the right-angled triangle "22.5=4.5*(4.5+x)" -> "x=0.5" so the answer is 5. I can use that trigonometric function because it's an inscribed triangle so it's an right-angled triangle too.
Interestingly, you can make the sliding thing rigorous. The key is that if you extend everything to a circle and rotate the yellow circle around the red circle, then consider the path traced by the rightmost point of the yellow circle at every step you'll get the black circle. Two steps to prove this. First show that the path traced by the rightmost point of the yellow circle is in fact a circle. Second, show that that circle is the same as the black circle. The second part is easier, so let's do that first. If the path is a circle, we can see that its diameter lies on the line containing the given diameter of the black circle and the leftmost point of the red circle will lie on both the path traced by the rightmost point of the yellow circle and is on the black circle. So as long as both circles have one more point in common they must be the same circle. That one additional point is the point given in the problem, namely the leftmost point of the yellow circle as given in the problem. Ok, now to prove that the path is in fact a circle. This is pretty straightforward by parametrizing the path. I'll use complex numbers to make the notation easier. Then the path is given by 1+2.5e^{i*theta}+1.5=2.5+2.5e^{i*theta}. Which is a circle of radius 2.5 with center at (2.5,0).
A bit simpler: you can show each semicircle and the chord are similar to each other. Draw the chord from the bottom left corner to the corner of the medium semicircle. That chord will pass through the two smaller semicircles at the single point they touch. You can conclude the angle of the chord relative to the base of all semicircles is the same. Then you can assume the ratio of the chord to any particular semicircle is also the same, lets call it x. So you get 2x + 3x = Dx and can conclude D is 5.
after watching every video you've posted the past year, I have come to the conclusion, to solve any geometry math problem, just make a bunch of triangles
All 3 circles are congruent, so 3 will always touch both shapes in this way if the 2 curves are touching, as long as both lines are parallel. Therefore congruent.
There is no calculation needed at all. Call small halfcircle C1, the bigger halfcircle C2 and the biggest C3. The reasoning goes as follows: Draw a line from the lower left vertex of C1 (A) to the touching point of C2 with C3 (B), then from there a line to the lower right vertex of C3 (C). As we know from Thales, △ABC is a right triangle. But the same argument goes for the smaller C1: Line from lower left vertex (A) to intersection of C1 with C2 (D), and from there to the lower right vertex of C1 (E). As one can see, the triangles △ABC and △ADE are similar, and one can draw a parallelogram then: E-C-B and upper left vertex of C2 (F). Therefore, the diameter of C2 (B-F = 3) is the same distance on the bottom (E-C = 3). Well, there *is* a calculation now: 2 + 3 = 5.
It may be something obvious, but how exactly did you conclude that the fourth vertex of the parallelogram with vertices E ,C, B is precisely the upper left vertex of C2 (F)?
@@RexIovis Given the points in my decription above: triangle △ABC is *similar* to triangle △ADE (both right triangled), so the sides BC and DE are parallel, and the extension of DE is towards F on C2, so BC = EF, because F is at the same height as B. So ECBF has to be a parallelogram.
Alternatively, after you found x=2, the distance on the bottom diameter from the far left corner to the foot of the vertical through the right top of the semicircle is 9/2. Now, instead draw a right triangle from that right top corner to the center of the big circle (on the horizontal line). This right has sides R (hypothenuse), 9/2-R and 3/2. It follows (9/2-R)^2+(3/2)^2=R^2. So R=10/4=5/2. Finally the diameter is 2R=5.
3:05 we can also use the property that h² = x1*x2 where h is the height from the hypotenuse to the right angle and x1 and x2 are the segment lengths of the hypotenuse after splitting it at the point of intersection with the segment h. We then get the following: 1.5² = 4.5 * y Rearranging for y: y = 2.25/4.5 = 0.5 4.5+0.5 = 5
I didn't assume the flat side of the yellow semi-circle was parallel to the flat side of the black semi-circunference, so that made things a bit confusing
I had a good time doing this one with analytic geometry. If you map the left end of the big circle's diameter to (0,0), the the intersection of the yellow circle and the big one ends up at (4.5,1.5). Then if you call the center of the big circle (c, 0), you get this nice equation from the circle formula: (4.5-c)^2+1.5^2=c^2 Great puzzle, thanks
I mean... I could tell because they're all half circles and each inner one is touching each other and one part of the outter one's curve. Because of that the outter one must have a diameter that is exactly the same as the sum of the two inner diameters. I doubt any of that is actual math logic, but it's the logic my brain used to solve it right away, and is also probably the reason why he started off showing that you could just visually move one over to see it.
The two inner circles presumably touch the outer circle at exactly one point, and touch each other at exactly one point. If that is the case, then there is a line that could be drawn through all three tangent points that would be both the sum of the diameters of the two inner circles as well as equal to the diameter of the larger circle. So it seems to me in these cases, you can always just sum the diameters of the two inner circles to get the diameter of the outer.
I used a little bit of analytic geometry, the first I said that the left bottom corner are (0,0) the Radius from the first are 1, and from the second 1.5, I know that the center from the first semi circle it's on (1,0) and the other on (x,1.5) but from the same method showed on the video I know that the difference of the coordinate x from the 2 centers is 2, so (1+2,1.5)=(3,1.5), and the point that the Big semicircle and the yellow semicircle share is a Radius form the yellow appart, so that point is (3+1.5,1.5)=(4.5,1.5) Now the equation from the Big semicircle are going to be y²+(x-r)²=r² So y= √(2xr-x²) and I need That the semicircle pass on the point (4.5,1.5) 1.5=√(2(4.5)r-(4.5)²) solving the equation, r=2.5, and the value from the answer it's 2r so 2(2.5)=5
My English it's not the Best, so sorry if I wrote something not good and if You tell I overcomlicate the problem using this method, I'm not so good with trigonometry
It took me a bit to get the answer (using similar methods to you), but after I did, I played around with it a bit and discovered a very simple and elegant way to solve this without trig or Pythagoras. Draw a line segment between the centers of the smallest and medium circles. It's super easy to see this line is 5/2 long (sum of the radii). Now, just slide the line segment along the edge of the largest half circle, until the end that was at the center of the medium half circle hits the edge of the largest half circle. The end that was at the center of the smallest circle started 1 from the edge of the largest circle and you slid it 1 radius of the medium half circle, i.e., 1.5, so it is now 5/2 from the edge of the largest circle. Since we know the center of the largest circle has to be on the bottom line, and we have two lines of the same length that go from a point on the bottom to the edge of the largest circle, this point must be the center point and the length of these lines must be the radius. We want two of these radii: 2*(5/2) = 5. Edit: I realized that while conceptually simple, it's wordy to explain. Seeing it visually, it is really very elegant.
@@Mqrius yeah, maybe it's a little buried in my explanation, but that is the key observation. You basically need 4 facts: 1) the two line segments terminate at points around the circle, 2) one of the segments points towards the center, 3) they are the same length, and 4) the other end of both terminates at the same point. This can only happen if these segments are radii.
This reminded me that there are many things we know to be true but cannot explain. We can quite intuitively see that the answer is 5, but proving it is 5 is a lot more challenging.
There's an assumption being made that hasn't been given. The yellow diameter drawn looks parallel to the diameter of red and the largest circle, and you used the assumption that it is when you made the 1.5 size square. But it's not a given that it's parallel, which could enlarge the large circle if the slope is positive, and vice versa if the slope is negative.
Catriona Agg's problems often aren't fully specified as an exam problem would be. They're more like physics problems in that you have to figure out what plausible assumptions you can make to get something you can solve (like lines that appear parallel being parallel, or quadrilaterals that look like squares being actually square) In fairness even the more traditional drawings for problems that mark parallels and right angles, often have the implicit assumption that curves are either circles or lines, and if three curves appear to intersect at a single point they actually do (both of which are also needed here)
If you mirror once at the middle of the yellow circle and once at the edge of the yellow circle, you see that the overlap of red and yellow in x is equal to the difference between the end of yellow and the end of black. q.e.d. geometrical proof
I'm not good at remembering the math but I instinctively look at problems like this and come up with the answer. I just understand some things by looking and used to get in trouble all the time because teachers and other students wouldn't believe me. It was always disappointing, especially because they wouldn't admit I was right after the answer was found
On the last steps, you can find y by using the geometric mean of the two segments formed on the hypotenuse. The geometric mean of 4.5 and y must equal the altitude which is 1.5. Solving for y, you can find that y = 1/2 or 0.5.
i found it out without math, Given that both lines are in the Larger halfcircle, we can deduce that if we where to grow one of the inner halfcircles and shrink the other at the same rate, until 1 is gone and the other is the same size, we can see that its a Proportional Relation and thus the lines of the straight have to be equal to the counter part
some back of the envelope math and wolfram alpha both say it's true. if you want to do some of the proof yourself. (x^2+2xy)^(1/2)+(y^2)/(x+y+(x^2+2xy)^1/2)=x+y have fun.
I had no way too prove it but my intuition told me that you can move the yellow one right corner along the black curve and it will always touch the red halve circle just as it does now, just in a different place
Uh! love this! Finding solutions to problems in creative ways is cool! Wish my teachers sold math as a puzzle instead of just giving them to me as work to be done.
If you want a quick intuitive idea: Think of the shapes as circles, and a circle as a path through all points at a given distance to some point. Now, what path does the black circle describe? It's the path of the yellow circle's right-most point if the yellow circle was to "roll around" the red one. This path must be identical to the black circle since the yellow circle's right point and the red circle's left point are both on it. So its diameter must be the sum of the red and yellow circles' diameters.
wen you really think about it, because its all circles. you can rotate the yellow circle around the red and just use there combined lenght. Its feels cheaty, but its just how circles work and this video proves it.
I was kind of a rocks-for-brains kid who couldn't stand the thought of doing math back in lower levels of school. Only now do I see the merit of math, after already completing one year of college. Whoops. This video was fantastic to watch.
A very beautiful problem, a clean and clear solution, even the graphics are perfect. Well... almost perfect. As a math teacher and graphic designer, I can ask: there is so much space on the screen, why is everything so small?? I watch the channel on the phone screen and I can say that the θ letter, in the tiny triangle on the right, towards the end of the solution, was 0.2 millimeters in size. Why???😮 In any case, a very fun and beautiful video. 👍
Now it is necessary to prove why smaller semicircles can fit into a large semicircle, the sum of the diameters of which is equal to the diameter of the largest semicircle
Very nice. You could even shorten the last part with the geometric mean theorem or right triangle altitude theorem (i dont know how it's called in english). For the triange it is true, that 1,5²=4,5•q. Very nice video! Thank you so much.
Good one! I tried to solve the general case but wound up in quartic territory with radicals all over the place. That might be a fun one to show us in a future video.
@@matthewrayner571 yeah me neither i was thinking about it a lot and it does get on my nerves that there us no straightforward explanation to it because it would be very nice
@@plibb I got one! Then I looked at the comment section again and saw it was there already, so I wasn't even the first. Very sad. Basically connect the centres of the medium and small semicircles as in the video, then slide that line to the right by the radius of the medium semicircle. You have two lines from a point on the big one's diameter to the edge, both of length equal to the sum of the radii of the small and medium semicircles. Therefore this point must be the centre, and the lines must be radii of the big semicircle.
I don't usually look for manipulations because of the whole 'Not drawn to scale' thing, but I was funny seeing that it worked out here. I can't tell if drawings like these are just inherently drawn to scale or not, so I just let the maths do the talking. Especially since visual proofs are just rarely accepted proofs anyway, if ever.
Doing the work to verify that it is 5 feels more rewarding. I was thinking of using a triangle at the start but got lost lol
Yeah, if you don't verify mathematically, you can't really know that what visually "looks right" is right. A good example of this is the disappearing chocolate piece when you slice a chocolate bar the right way.
@@HenrikMyrhaug At the end how does he know that the big triangle and the small triangle both have the same alpha angle?
This is because we saw how the biggest triangle(big+small) has alpha and theta angle along with a right angle. And we know the sum of angles in a triamgle is 180. So we get
Alpha + theta + 90(right angle) =180
Since we know both triangles have a right angle and theta angle, we can concludw that both triangles contain the alpha angle.
Hoped this made sense
@@tcjgaming9813 tysm bro
@@mrmailman5561not necessarily the same he’s using theta to show those particular angles and the rest is all mix of “sohcahtoa” and other ratios between those angles and their relationship to the circle they are a part of.
Edit: aaaaaaaaaaaaand I just saw you got the answer already lol sry I’m hi
It IS exactly 5. How exciting!
Happy 250k subs! How exciting!
let's put a box around it! how exciting!
@@mohammedyuan how exciting!
I'm so happy I got part of the solution correct just by eye-balling it, I watched enough of your videos to instinctively connect the 2 lines of the circles from the center and I also saw the small right triangle with the Y section, your videos really are helpful in helping with problem solving, Andy, thanks a lot
Same! Watching enough of the visualizations in the videos really helps to conceptualize the steps through it
I fricking love your "how exciting" endings lol
0:20 They don't let us do it in school.
Ye you shouldnt be allowed image can be missleading
well yeah... that's a nice way to verify the approximate answer but not more
i remember seeing the similar video where it was quite obvious to me that the answer was 3,5 based on a grid. but it turned out to be something like 1,23π
you really never know...
Quick solution using homothety:
Let the red and yellow semicircles be tangent at T and touch the black semicircle at A and B, respectively. Let |AT| = x. Consider the homothety from T (with factor -1.5) sending A to B and notice it maps the red semicircle to the yellow one => A, B and T are on a line and |TB| = 1.5x, so |AB| = x + 1.5x = 2.5x. Now consider the homothety from A sending T to B and notice it maps the red semicircle to the black semicircle, meaning the ratio of their diameters is the ratio of chords AT and AB, namely 2.5x / x = 2.5 hence the unknown radius is 2.5 * 2 = 5.
I usually think through problems like these, but I just imagined sliding the yellow semi circle down and went, “It’s five”
Oh, I just watched a bit more and that’s what he did
Inspiring
It would be interesting if you showed if this diameter being the sum of both diameters is a coincidence here or true for any combination of two diameters.
I did the math using arbitrary diameters and all complicated math gets cancelled out until there is only the sum of the diameters! Pretty much the same way he did it, just using a,b instead of 2,3.
I showed that it is true! See my first comment here.
It is indeed true for any combination of diameters.
Assume the two smaller semi-circles have radii a and b (diameters 2a and 2b).
We can show that larger semi-circle has diameter 2a + 2b.
We'll divide the diameter into segments of length a, c, b, and y (see diagram at 2:45.
Since c is leg of right triangle, whose other leg = b, and hypotenuse = a+b, we get
c² + b² = (a+b)²
c² = a² + 2ab
Now we calculate y using similar triangles, as shown in video:
y/b = b/(a+c+b)
y = b²/(a+b+c)
Now we calculate diameter of larger semi-circle:
d = a + c + b + y
d = a + b + c + b²/(a+b+c)
d = ((a+b+c)² + b²) / (a+b+c)
d = ((a² + b² + c² + 2ab + 2ac + 2bc) + b²) / (a+b+c)
Replace c² with a² + 2ab
d = (a² + b² + (a² + 2ab) + 2ab + 2ac + 2bc + b²) / (a+b+c)
d = (2a² + 2b² + 4ab + 2ac + 2bc) / (a+b+c)
d = (2 (a²+b²+2ab) + 2c(a+b)) / (a+b+c)
d = (2 (a+b)² + 2c(a+b)) / (a+b+c)
d = 2 (a+b) (a+b+c) / (a+b+c)
d = 2 (a+b)
d = 2a + 2b = (sum of diameters of smaller semi-circles)
Awesome solution with similar triangles!
How I did it:
Equation of the circle is (x-h)² + y² = r².
Circle goes through (0,0) so plug in x=0 and y=0 to get h = r. Thus the equation is
(x-r)² + y² = r².
Using the same reasoning as you did to find the 3-4-5 triangle, the coordinates of the bottom of the yellow semicircle are (3,0), and thus the rightmost point is (4.5,1.5).
Plug in x=4.5, y=1.5 into the equation: (4.5-r)² + 1.5² = r². This becomes an easy linear equation, and r=2.5. Thus the diameter d = 2(2.5) = 5.
This way feels more down and dirty compared to your elegant solution!
I love how you always knew it was 5 but you really took the time to teach us the geometry rules that prove its 5 and how we know that. its really finding complex structures within simple questions that does it for me thanks andy
Math is beautiful
Amazing work! Thank you for clearly explaining each step and not skipping anything.
At the end how does he know that the big triangle and the small triangle both have the same alpha angle?
@@mrmailman5561 They are the literal same alpha angle. The small triangle is a portion of the big triangle. 3:24
At 2:41 you could have used intersecting chord properties i.e. y x 4.5 = 1.5 x 1.5 , imagining the rest of the circle, so y =0.5.
Your initial intuition inspired another solution: Suppose you extend both the yellow and orange semi-circles to be complete circles. Now imagine the yellow circle orbiting clockwise around the orange circle, keeping tangent. During that orbit the distance between their centers obviously remains constant, namely 2.5 units (sum of the radii). The sum of their diameters is just double that, namely 5 units. When the yellow circle's center is on the bottom line of the large semicircle, it's still 5 units, so 5 must be the diameter of the large semicircle.
Interesting thinking but this is not a full solution. After the yellow circle is rotated around the smaller one so its center lies on the bottom line, how do you know its right edge lines up perfectly with the corner of the larger semicircle?
I ended up using “perpendicular bisector to any chord passes through the center of the circle” and drew bisector on the hypotenuse of 4.5 and 1.5. Your method is simpler though.
There's another (2,1.5,2.5) rightwing triangle between the center of the black circle, the rightmost point where yelllow circle touches the black circle and the point on the line that is distance y from the right. From this other triangle together with the first one, we can derive the diameter of the large circle.
I have to admit that the prob is really fun. I did it using the trigonometric function and Pythagoras theorem.
My solution is similar to the video, except the last step. After found the 4.5 side, I use Pythagoras theorem to find the hypotenuse^2 of the big triangle "4.5^2+1.5^2=22.5". Followed by the trigonometric function in the right-angled triangle "22.5=4.5*(4.5+x)" -> "x=0.5" so the answer is 5. I can use that trigonometric function because it's an inscribed triangle so it's an right-angled triangle too.
Interestingly, you can make the sliding thing rigorous. The key is that if you extend everything to a circle and rotate the yellow circle around the red circle, then consider the path traced by the rightmost point of the yellow circle at every step you'll get the black circle.
Two steps to prove this. First show that the path traced by the rightmost point of the yellow circle is in fact a circle. Second, show that that circle is the same as the black circle.
The second part is easier, so let's do that first. If the path is a circle, we can see that its diameter lies on the line containing the given diameter of the black circle and the leftmost point of the red circle will lie on both the path traced by the rightmost point of the yellow circle and is on the black circle. So as long as both circles have one more point in common they must be the same circle. That one additional point is the point given in the problem, namely the leftmost point of the yellow circle as given in the problem.
Ok, now to prove that the path is in fact a circle. This is pretty straightforward by parametrizing the path. I'll use complex numbers to make the notation easier. Then the path is given by 1+2.5e^{i*theta}+1.5=2.5+2.5e^{i*theta}.
Which is a circle of radius 2.5 with center at (2.5,0).
A bit simpler: you can show each semicircle and the chord are similar to each other.
Draw the chord from the bottom left corner to the corner of the medium semicircle. That chord will pass through the two smaller semicircles at the single point they touch.
You can conclude the angle of the chord relative to the base of all semicircles is the same. Then you can assume the ratio of the chord to any particular semicircle is also the same, lets call it x. So you get 2x + 3x = Dx and can conclude D is 5.
you can also get y through intersecting chords theorem y(4.5) = (1.5)(1.5)
cancel 1.5 both sides: 3y = 1.5
y = 0.5
after watching every video you've posted the past year, I have come to the conclusion, to solve any geometry math problem, just make a bunch of triangles
that's evident.
any polygon can be made with triangles so you only need triangles to make any shape.
ok. maybe you need a bit of trigonometry.
All 3 circles are congruent, so 3 will always touch both shapes in this way if the 2 curves are touching, as long as both lines are parallel. Therefore congruent.
There is no calculation needed at all. Call small halfcircle C1, the bigger halfcircle C2 and the biggest C3.
The reasoning goes as follows:
Draw a line from the lower left vertex of C1 (A) to the touching point of C2 with C3 (B), then from there a line to the lower right vertex of C3 (C). As we know from Thales, △ABC is a right triangle. But the same argument goes for the smaller C1: Line from lower left vertex (A) to intersection of C1 with C2 (D), and from there to the lower right vertex of C1 (E). As one can see, the triangles △ABC and △ADE are similar, and one can draw a parallelogram then: E-C-B and upper left vertex of C2 (F). Therefore, the diameter of C2 (B-F = 3) is the same distance on the bottom (E-C = 3).
Well, there *is* a calculation now: 2 + 3 = 5.
It may be something obvious, but how exactly did you conclude that the fourth vertex of the parallelogram with vertices E ,C, B is precisely the upper left vertex of C2 (F)?
@@RexIovis Given the points in my decription above: triangle △ABC is *similar* to triangle △ADE (both right triangled), so the sides BC and DE are parallel, and the extension of DE is towards F on C2, so BC = EF, because F is at the same height as B. So ECBF has to be a parallelogram.
@@hcgreier6037 By "extension of DE is towards F on C2" do you mean to say that E, D, F are collinear? How do you know that?
@@RexIovis I know that from Mr. Thales: C1: △ADE => right angle at D, C2: △BDF => right angle at D
@@hcgreier6037 Ah, thank you, this is what I was missing. Nice solution.
Alternatively, after you found x=2, the distance on the bottom diameter from the far left corner to the foot of the vertical through the right top of the semicircle is 9/2. Now, instead draw a right triangle from that right top corner to the center of the big circle (on the horizontal line). This right has sides R (hypothenuse), 9/2-R and 3/2. It follows (9/2-R)^2+(3/2)^2=R^2. So R=10/4=5/2. Finally the diameter is 2R=5.
3:05 we can also use the property that h² = x1*x2 where h is the height from the hypotenuse to the right angle and x1 and x2 are the segment lengths of the hypotenuse after splitting it at the point of intersection with the segment h. We then get the following:
1.5² = 4.5 * y
Rearranging for y:
y = 2.25/4.5 = 0.5
4.5+0.5 = 5
I didn't assume the flat side of the yellow semi-circle was parallel to the flat side of the black semi-circunference, so that made things a bit confusing
I think without that bit it's unsolvable
You're correct. That information is not given, and without it there is no way to solve the problem.
I had a good time doing this one with analytic geometry. If you map the left end of the big circle's diameter to (0,0), the the intersection of the yellow circle and the big one ends up at (4.5,1.5). Then if you call the center of the big circle (c, 0), you get this nice equation from the circle formula:
(4.5-c)^2+1.5^2=c^2
Great puzzle, thanks
Wow! Great (and fun) math challenge. Thanks for your channel.
I mean... I could tell because they're all half circles and each inner one is touching each other and one part of the outter one's curve. Because of that the outter one must have a diameter that is exactly the same as the sum of the two inner diameters.
I doubt any of that is actual math logic, but it's the logic my brain used to solve it right away, and is also probably the reason why he started off showing that you could just visually move one over to see it.
The two inner circles presumably touch the outer circle at exactly one point, and touch each other at exactly one point. If that is the case, then there is a line that could be drawn through all three tangent points that would be both the sum of the diameters of the two inner circles as well as equal to the diameter of the larger circle. So it seems to me in these cases, you can always just sum the diameters of the two inner circles to get the diameter of the outer.
I love you Andy! Always have the sweetest possible questions! classic and simple but yet not too easy!
I used a little bit of analytic geometry, the first I said that the left bottom corner are (0,0) the Radius from the first are 1, and from the second 1.5, I know that the center from the first semi circle it's on (1,0) and the other on (x,1.5) but from the same method showed on the video I know that the difference of the coordinate x from the 2 centers is 2, so (1+2,1.5)=(3,1.5), and the point that the Big semicircle and the yellow semicircle share is a Radius form the yellow appart, so that point is (3+1.5,1.5)=(4.5,1.5)
Now the equation from the Big semicircle are going to be y²+(x-r)²=r²
So y= √(2xr-x²) and I need That the semicircle pass on the point (4.5,1.5)
1.5=√(2(4.5)r-(4.5)²) solving the equation, r=2.5, and the value from the answer it's 2r so 2(2.5)=5
My English it's not the Best, so sorry if I wrote something not good and if You tell I overcomlicate the problem using this method, I'm not so good with trigonometry
It took me a bit to get the answer (using similar methods to you), but after I did, I played around with it a bit and discovered a very simple and elegant way to solve this without trig or Pythagoras. Draw a line segment between the centers of the smallest and medium circles. It's super easy to see this line is 5/2 long (sum of the radii). Now, just slide the line segment along the edge of the largest half circle, until the end that was at the center of the medium half circle hits the edge of the largest half circle. The end that was at the center of the smallest circle started 1 from the edge of the largest circle and you slid it 1 radius of the medium half circle, i.e., 1.5, so it is now 5/2 from the edge of the largest circle. Since we know the center of the largest circle has to be on the bottom line, and we have two lines of the same length that go from a point on the bottom to the edge of the largest circle, this point must be the center point and the length of these lines must be the radius. We want two of these radii: 2*(5/2) = 5.
Edit: I realized that while conceptually simple, it's wordy to explain. Seeing it visually, it is really very elegant.
How do you know that the line you moved is tangent to the circumference of the big circle?
@@Mqrius I don't. It doesn't need to be.
@@atrus3823 Oh right I missed the bit about it being equal length as the horizontal line. That's pretty neat
@@Mqrius yeah, maybe it's a little buried in my explanation, but that is the key observation. You basically need 4 facts: 1) the two line segments terminate at points around the circle,
2) one of the segments points towards the center, 3) they are the same length, and 4) the other end of both terminates at the same point. This can only happen if these segments are radii.
This reminded me that there are many things we know to be true but cannot explain. We can quite intuitively see that the answer is 5, but proving it is 5 is a lot more challenging.
There's an assumption being made that hasn't been given. The yellow diameter drawn looks parallel to the diameter of red and the largest circle, and you used the assumption that it is when you made the 1.5 size square. But it's not a given that it's parallel, which could enlarge the large circle if the slope is positive, and vice versa if the slope is negative.
Catriona Agg's problems often aren't fully specified as an exam problem would be. They're more like physics problems in that you have to figure out what plausible assumptions you can make to get something you can solve (like lines that appear parallel being parallel, or quadrilaterals that look like squares being actually square)
In fairness even the more traditional drawings for problems that mark parallels and right angles, often have the implicit assumption that curves are either circles or lines, and if three curves appear to intersect at a single point they actually do (both of which are also needed here)
If you mirror once at the middle of the yellow circle and once at the edge of the yellow circle, you see that the overlap of red and yellow in x is equal to the difference between the end of yellow and the end of black. q.e.d. geometrical proof
I'm not good at remembering the math but I instinctively look at problems like this and come up with the answer. I just understand some things by looking and used to get in trouble all the time because teachers and other students wouldn't believe me. It was always disappointing, especially because they wouldn't admit I was right after the answer was found
On the last steps, you can find y by using the geometric mean of the two segments formed on the hypotenuse. The geometric mean of 4.5 and y must equal the altitude which is 1.5. Solving for y, you can find that y = 1/2 or 0.5.
i found it out without math, Given that both lines are in the Larger halfcircle, we can deduce that if we where to grow one of the inner halfcircles and shrink the other at the same rate, until 1 is gone and the other is the same size, we can see that its a Proportional Relation and thus the lines of the straight have to be equal to the counter part
1:37 i was kinda interested, now im intrigued
0:25 so the answer was 5, if we dragged the red down it would of shown us that we were correct
i loved this. with my high-school education, i was able to follow everything, only not recognizing 2 things you mentioned.
Enjoyed this one very much! How exciting indeed 😊
[Need to proof] I guess any 2 semi-cycles in a bigger semi-cycle created this way gonna have the same property.
some back of the envelope math and wolfram alpha both say it's true. if you want to do some of the proof yourself.
(x^2+2xy)^(1/2)+(y^2)/(x+y+(x^2+2xy)^1/2)=x+y have fun.
I had no way too prove it but my intuition told me that you can move the yellow one right corner along the black curve and it will always touch the red halve circle just as it does now, just in a different place
Uh! love this!
Finding solutions to problems in creative ways is cool!
Wish my teachers sold math as a puzzle instead of just giving them to me as work to be done.
If you want a quick intuitive idea: Think of the shapes as circles, and a circle as a path through all points at a given distance to some point. Now, what path does the black circle describe? It's the path of the yellow circle's right-most point if the yellow circle was to "roll around" the red one. This path must be identical to the black circle since the yellow circle's right point and the red circle's left point are both on it. So its diameter must be the sum of the red and yellow circles' diameters.
Did we establish that the base of the yellow semicircle was parallel to the base of the big semicircle? Does it matter?
The radius of a semicircle would stay the same no matter the angle, so I don't think it matters
Andy, I love your videos. Thanks for the excellent visuals, clear breakdowns, and passion for math.
May I ask: what’s your educational background?
Solved it after seeing the thumbnail. Feels good to see I wasn't wrong with my answer.
wen you really think about it, because its all circles.
you can rotate the yellow circle around the red and just use there combined lenght.
Its feels cheaty, but its just how circles work and this video proves it.
getting back into 6th grade maths thanks to you, daaamn
I also found 5 but without doing anything. Except calculating 3+2 🤣
I had the same idea, but didn't think of similar triangles
I've seen this question before and it never solved it fully. Great explanation! It makes sense now. If I see it again I'll **** *** **** out of it.
01:42 can just multiply all sides by 2. Then we see that is 3 4 5 triangle. So 2x = 4...
I was kind of a rocks-for-brains kid who couldn't stand the thought of doing math back in lower levels of school. Only now do I see the merit of math, after already completing one year of college. Whoops. This video was fantastic to watch.
Pretty satisfying one, visually simple and the solution isn't terribly complicated you just gotta work your way from right to left in a cool way.
I just watch these for the dopamine rush I get when you say "how exciting."
andy you are my spirit animal. congrats on 250k
A very beautiful problem, a clean and clear solution, even the graphics are perfect.
Well... almost perfect.
As a math teacher and graphic designer, I can ask: there is so much space on the screen, why is everything so small??
I watch the channel on the phone screen and I can say that the θ letter, in the tiny triangle on the right, towards the end of the solution, was 0.2 millimeters in size. Why???😮
In any case, a very fun and beautiful video. 👍
Such a relief. I thought it was gonna be 5.14 and so on
Now it is necessary to prove why smaller semicircles can fit into a large semicircle, the sum of the diameters of which is equal to the diameter of the largest semicircle
the smile fading after he said how exciting
Bro is better at teaching math than my teacher
Instead of saying 5 I'd say that the yellow semicircle can be inverted and to perfectly fit with the red semicircle on the diameter
step 1: ruler
Finally a math question i can do.
Very nice. You could even shorten the last part with the geometric mean theorem or right triangle altitude theorem (i dont know how it's called in english). For the triange it is true, that 1,5²=4,5•q.
Very nice video! Thank you so much.
Spectacular work!
Dude, that was obviously a 5-4-3 triangle at the beginning.
(4.5-r/2)^2+1.5^2=(r/2)^2
20.25-4.5r+2.25=0
r=(20.25+2.25)/4.5=5
Wow, I wasn't expected the last step
Great video! Very well done.
Oh man this is a machine problem like a 1st grade in high school in Poland 😂
Good one! I tried to solve the general case but wound up in quartic territory with radicals all over the place. That might be a fun one to show us in a future video.
Math is life and life is math.
I love geometry. Do these type of questions only please ❤
What a easy question! In reality it's the easiest question what I have done today.
Thats a great explanation Andy.
Always gotta work through it if they don't say everything was drawn to proportion.
I did math for a long time ago. But it took 5-10 seconds to see the solution
You cannot approach these problems with image manipulation. Most are not made to scale so math is the best approach to these questions.
Anyone else think Andy is cute?
Normal speed, how exciting!
My solution is exactly like yours
How exciting!
The triangle notion explanation was that clear, thanks indeed!
I’ve literally learned more by watching this video then during 12 years of elementary and middle school math classes.
Nice! Really like your content and not hearing cancel. Keep them coming!
What app or software do you use for explaining and solving these problems ?
because it looks very amazing ! 👍🏻
Math really is fun when theres no time pressure and school grades involve 😢
Cool. Now do the proof that if two kegs of a triangle inscribed in a circle are on the diameter then it’s a right triangle?
i love the fact that you did this and didnt think "what if i moved the big circle down since i know the combined radii is 5"
It's not straight down, though. It's down and to the right. And I can't yet find an elegant geometric proof that shows it slots into place.
@@matthewrayner571 yeah me neither i was thinking about it a lot and it does get on my nerves that there us no straightforward explanation to it because it would be very nice
@@plibb I got one! Then I looked at the comment section again and saw it was there already, so I wasn't even the first. Very sad.
Basically connect the centres of the medium and small semicircles as in the video, then slide that line to the right by the radius of the medium semicircle. You have two lines from a point on the big one's diameter to the edge, both of length equal to the sum of the radii of the small and medium semicircles. Therefore this point must be the centre, and the lines must be radii of the big semicircle.
@@matthewrayner571 oh yeah, thanks
A great challenge for anyone in geometry class
Semi circles and triangles, how exciting!
Nice but I did this in like 3 seconds just eyeballed and said 5. Saved all the time
How exciting indeed! 😊
Awesome, very easy to understand explanations :)
I don't usually look for manipulations because of the whole 'Not drawn to scale' thing, but I was funny seeing that it worked out here. I can't tell if drawings like these are just inherently drawn to scale or not, so I just let the maths do the talking. Especially since visual proofs are just rarely accepted proofs anyway, if ever.
Can you do it with calculus. By proving they the surface of the sphere staying tangent they trace a path making it 5