Nice to have a little puzzle like this that's just algebra once in a while. :)

Wonderfully explained, and perfectly concise. I clicked on this out of pure interest, but if I was trying to cheat on my math homework, this would’ve been a great guide, and that’s the highest honor I can bestow.

This awesome and all but can we appreciate how straight he draws the lines

At 2:44, note that the two x's are the same by virtue of the fact that two tangent lines to a circle from a single external point must be equal. The same logic applies to the two y's. This can be written down without any additional explanation.

im japanese and i can confirm school math life was hell for everyone.

i forgot how beautiful high school algebra could be once you had all the tools required. Just plug stuff in in a way that makes sense and presto

lol, i "cheated" by using the half-angle formula for tan. with that you can easily find that x=2r and y=3r and the rest is trivial. with that you can also generalize the problem for an arbitrary right-angled triangle with sides a,b,c, in that case you have r = abc/[(a+b+c)(a+b)]

I loved seeing the relationships drawn. After that, I felt tempted to draw a coefficient matrix and solve for r using cramer's method.

I always loved constructing and solving for variables like this in highschool geometry. It's basically just puzzle solving with math!

Thank you for these videos. I look at one each day at work. Very interesting and brain excercising.

This is such a nice way to start the brain in the beginning of the day... You talk and demonstrate like some of my favourite teachers. I've not learn/revise anything since I left school though always enjoyed maths and geometries.

You can also extend the line segment connecting the centers of the circles until it meets the sides of length 3 and 4, then use the 3 new triangles that creates (all similar to the big triangle), then use those triangles to find what you called x and y in terms of r without ever actually setting up a system of equations.

Excellent video!! Thank you!

I think geometry is quite difficult as I’m still learning this side of math in school. This has opened some interesting ways of thinking for me.

6:46

So cool. Excellent explanation! Thanks.

You're a really great teacher

that was really good, concise and easy to follow

I still enjoy how he ends with "and that's a good place to stop" and abruptly ends the video. It allows the viewer to wake up from the problem and math itself.

I'm not a math freak at all, but this is undoubtedly beautiful.

At ~2:00, the two triangles are similar, so the inner triangle's vertical is 6r/5. Extend the blue vertical to meet the outer triangle, forming a new similar triangle, with height 5r/4. From that new intersection, draw a horizontal line left to meet the outer triangle, forming a new similar triangle, with height 3r/4. Add them up, so the vertical of the outer triangle is (3/4 + 5/4 + 6/5 + 1)r = 3 => r = 5/7. This is a bit simpler. Do the wood-block prints come with solutions?

Love these sangaku videos

Your explanation is very clear.

Very elegant! I love it.

nice job, mike

I wrote the hypotenuse as -3x/4 - y + 3 = 0 and the centre of the leftmost circle as (r, 11r/5) and used the point-from-line distance formula.

Very satisfying and very well explained

thats really great

This is a way better way of demonstrating system of equation technique than anything we learned in algebra class, thank you

Mindblowing!

Good one !

This dude has mastered the art op stopping: 'That was our final goal, and that is what we done, and that is a good place to stop'. Perfect.

I was wondering whether it was possibile to show that, given a right triangle the radius of the two inner circles is equal to the ipothenhse over the sum of the catheta, since 5 /7 = 5/(3+4)

Extend the left side of the triangle downwards. Draw a perpendicular bisector of the segment joining the centers of the circles. Two congruent triangles will be created (they have the same circles inscribed), which are similar to the large triangle. Their sides have an aspect ratio of 3:4 and add up to 5. So they have sides 3*5/7 = 15/7 and 4*5/7 = 20/7. The third side of these triangles is 5*5/7 = 25/7. Therefore the radius of the inscribed circle (from the formula for the radius of the circle inscribed in a right triangle) = (15/7 + 20/7 - 25/7) / 2 = 5/7.

This was really fun.

I was always wondering this

That was beautiful

I took a random guess at the beginning of the video of 2/3 (more accurately a diameter of 4/3 first) close! Always nice to see the numbers work out.

Cool shapes

This is so dope

Wow simply brilliant. 👏👍😊

Beautiful

This channel feels like a massive flex

For general a,b c we have r = a*(b+c-a)/2*(b+c)

Give me a million years and I still wouldn't have thought of drawing 4 radii and rectangles etc. Brilliant.

This is the formula that works for *any* triangle and n touching circles. 1/r = (2n-2)/c + (a+b+c)/(2A), where r is the radius of the n circles touching side c, and A is the area of the general triangle. In this particular case n=2 and A=6, so 1/r=2/5+1 and r=5/7. Deriving the formula is not too difficult, but it naturally uses some other triangles than Michael introduced here.

Feels good to go back through geometric problem solving.

I calculated this using the bissectrice of both angles and thus finding an expression for x=r/(tan(0.5*atan(4/3)) = 2r and y=3r (I coincedently used the same letters for the same lines) and then finally 2r+2r+3r=5 -> r=5/7

Seems the general case for a right triangle with legs a and b (a

There are even more beautiful substitutions: v=x+r and w=y+r. So the system of equations is: 5=v+w; 4/5=(4-w)/2r; 3/5=(3-v)/2r

This is a good solution, i almost didn't think of it. Of course, it relies on the fact that the drawing is in someway drawn to its scale rather than drawn to resemble the scale.

Decompose into 3 triangles and one trapezoid. Each has area depends on r but altogether has to have area 6. This can solve for r as well (in fact, it's very easier). [1/2•3•r + 1/2•4•r + 1/2•(12/5-r)r + 1/2 (5+2r)•r = 6]

Thank you..

did it a *bit* of a different way that doesn't involve triple simultaneous equations: -let the radii = x (I know they used r but x is easier since no other unknowns are used) -the distance between both centres of circles is 2x -the radii to the points where the circles meet the hypotenuse makes 2 right angles since a tangent (the hypotenuse) is perpendicular to a radius -two opposite lengths are the same and there are 2 90º angles so the quadrilateral between the 2 radii, the length between both centres and the length between both points where the circles meet the hypotenuse is a rectangle, so the length between points where the circles meet the hypotenuse is 2x -if we draw a radii of the left circle that is parallel to the line length 3 and a radii of the right circle that is parallel to the line length 4 then we form a right angled triangle -we already know the two lengths of said right angled triangle are parallel to the outer triangle's two shorter lengths and also the hypotenuse to the small right angled triangle is parallel to the bigger hypotenuse so the smaller right angled triangle is similar to the bigger one -the scale factor from the big right angled triangle to the small one is 2x/5 based off of each one's hypotenuses -we can use this to find that the other lengths of the small right angled triangle are 8x/5 and 6x/5 -we can now draw a radii of each circle that meets the two smaller sides of the big right angled triangle -now we can find that the lengths from the bottom left point of the big right angled triangle to the points where the newly constructed radii intersect with it are x + 8x/5 and x + 6x/5 respectively -this means the rest of each lengths are 3 - x - 6x/5 and 4 - x - 8x/5 -tangents to a point are equal so we can find the equivalent of X and Y from 2:18 in the video -we now know that 3 - x - 6x/5 + 4 - x - 8x/5 + 2x = 5 -solve for x to get 5/7 -to generalise, let 3, 4 and 5 be replaced with a, b and c respectively -we get that a - x - 2ax/c + b - x - 2bx/c + 2x = c -solve for x to get c(a+b-c)/2(a+b) looks like I rambled on a bit so TL/DR of the method: prove that the inner right angled triangle between the two circles' centres is similar to the bigger one, find the length of the smaller lengths in terms of the radius, find the length of the hypotenuse in terms of the radius, solve for the radius.

Circle A: (x-r)(x-r) + (y-h)(y-h) = rr Circle B: (x-k)(x-k) + (y-r)(y-r) = rr The circles are tangent to the hypotenuse and are of equal radii Therefore the line connecting the centers must be parallel to the hypotenuse. And so the h-r:k-r:2r triangle must be in a 3:4:5 ratio... k = 13r/5 h = 11r/5 Hypotenuse: y = -3x/4 + 3 Substituting this, and h, back into Circle A gives us the intersection x at the hypotenuse in terms of r: (x-r)(x-r) + (-3x/4+(15-11r)/5)(-3x/4+(15-11r)/5) = rr To simplify, let Q = (15-11r)/5 [25/16]xx - [(60+13Q)/22]x + [QQ] = 0 As the intersection is a tangent, the discriminant of the quadratic formula is zero... Giving us a quadratic in Q: [2856]QQ + [-1560]Q + [-3600] = 0 Q = 10/7 OR Q = -15/17 r = (15-5Q)/11 r = 5/7 (inside the triangle) *The solution* OR r = 30/17 (outside the triangle)

how can I watch this video and suddenly grasp what 3 years of high school teachers couldn't drone into my head over and over. good job you did what all my math teachers failed to do through the years, let me grasp what was going on.

If you see closely fro any perpendicular triangle u can divide the hypotenuse with the sum of the adjacent sides

"And that's a good place to stop" this guy can write a novel

have no idea why this was recommended but i enjoyed it, thank you

Here are some follow up questions that came to my mind: Level 1: Find the general formula for 'r' for any right-angled triangle with sides a, b, & c = sqrt(a²+b²). Level 2: Say in any right-angled triangle, the two circles were not similar and could be shifted in size, then find the maximum value of R+r (in terms of a,b, & c), where R & r are the radii of the two circles in question. Level 3: Find r if there are 3 similar triangles instead of just 2 in the base question. Then generalise the result for any right-angled triangle. Level 4: Find the maximum sum of the three radii if the circles are not fixed to be similar. Level 5: Prove the Riemann Hypothesis. xD

So in a question like this regardless of the length of the sides of the triangle, will the radius always be hyp/adj+opp or was it just a coincidence this time?

Good explanation, just clicked for fun and stay 😂

Watching him at 1:20 was a real, '''''wow'''''' moment! I didn't know where to go from there anymore than I knew from the start but that was so freakin' pretty to watch!!

... Or, equivalently, taking the three equations that he boxed in at 5:05, if we add the lower two equations we obtain 24r + 5(x+y) = 35 then, with our eyes on the remaining equation, above we write out 14r + 5(2r + x+y) = 5(5) + 10 Subtracting off 5 times top equation, we are left with 14r = 10 ==>. r = 5/7 Equivalent, but just a little quicker, and maybe a little less brute force.

That's a great puzzle, I will rate it 5/7

I'm mathematically challenged, how you figured that out was amazing

Instant sub.

Let a be the height, b the base, c the hypotenuse and (P , R) the center of the right circle with radius R. Define T = a + b + c and S = a + b - c. Then TS = 2ab. We have (ab - aP - bR) / c = R , where the left side is the distance of (P , R) to the line of equation ax + by = ab. So P - R = (ab - RT) / a = 2R * (b / c). (by similarity of triangles) Rearranging, we have abc = 2abR + RTc = RT * (S + c) So R = (a * b * c) / ( (a + b + c) * (a + b) )

One can show that for any right angle triangle the radius r of the two smaller circles is related to the radius of the single inscribed circle R by the nice formula r = R * c / (a+b). Next step, generelize to n smalle circles?

Very nice but there is no need to add variables x and y without adding ant more math. From the moment he draws the blue lines and creates the internal triangle, ona can use proportionality of the two (external and internal) triangles to say: 3 : 5 = "the middle segment of the vertical side" : 2r --> "the middle segment of the vertical side" = 6r/5 4 : 5 = "the middle segment of the horizontal side" : 2r --> "the middle segment of the vertical side" = 8r/5 Notice that he uses this property as well, so I am not adding more math here. This way, we immediately get that x = 3 - 11r/5 y= 4 - 13r/5 The advantage is that we do not need to carry around x and y and talk about a system of 3 equations in 3 variables, but we can directly jump to 3 - 11r/5 + 2r + 4 - 13r/5 = 5 and solve that r = 5/7 (this is my own solution before watching the video)

Goddamn, I had forgotten how much I loved math.

Nice ✨✨

Interesting that the radius is H/(L1+L2)

6:34 perfect

using half angle tangent formulas and using equation for the length of the segments adding up to 5, I get a roundabout way to the answer. a bit long, but satisfying nonetheless

That is so beautiful gonna make me shed a teat

Cool how it ends up being hypotenuse/(base+height) I wonder if that’s true for other triangles?

I did it differently. Y=r/tg(acos(4/5)/2) , X=r/tg(acos(3/5)/2) and X+2r+Y=5. I got 0.714. Its not an elegant solution by any means and heavily dependent on a calculator but it gets the job done

I don't know why I'm trying to comprehend this directly after waking up from a nap I didn't plan on taking. This is still very cool, and I would have went to Harvard if this guy was my math teacher.

I found the dimensions of the smaller triangle to be 1.2r, 1.6r, 2r To get the system of three equations: x + 1.2r + r = 3, y + 1.6r + r = 4, and x + y + 2r = 5. Solving the first two for x and y and substituting gives (3 - 2.2r) + (4 - 2.6r) + 2r = 5, which reduces to 2 = 2.8r or r = 5/7.

I got a tape measure if you need to burrow it might be a little quicker

let X as a distance from left bottom corner to circle touch. Y as a distance from right botton to touch circle. We have 3 equas: 1) X+6R/5 + R = 3 2) Y+8R/5 + R = 4 3) X+Y+2R = 5 1+2 - 3: 14R/5 = 7-5 => R = 5/7

The theorem Michael quoted about tangents to a circle from outside the circle re x and y, is called "the ice-cream cone theorem"

It can also be solved by taking semi angles of both acute vertices and dividing hypotenuse of bigger triangle into three parts all in the term of 'r'. Upper part comes as 2r lower part comes as 3r so 2r+3r+2r=5 Then r=5/7. Its so simple no need to solve this much of algebra just simple trigonometry.

Using the same method but keeping the sides of the triangle general (a, b, c), the formula becomes: r = 1/2 * c * (a + b - c) / (a + b).

I tried to do it without the 3, 4 and 5 values, because of that, I also tried to avoid x and y since I wouldn't get values for them, I ended up stuck... so I watched the resolution

How did you find the centerpoints though?

shapes in math. the bane of my existence

Great explanation! What grade level do you think this exercise is for?

For a 3:4:5 ratio triangle... R = Hypoteuse / ( Height + Base) R = 5 /( 3 + 4 ) = 5/7

Me, an intellectual: *Slaps a ruler on the radius* Ok but in all seriousness, it actually is possible to solve this geometry problem with a ruler (or tape measured if you need more length), hear me out. Now I know what you’re thinking, “We don’t even know what unit of measurement is being used!” And you’d be right. But the ruler can be used to find proportion. Just measure any side, let’s say the hypotenuse for example. It then measured 21 inches let’s say. Then we measure the radius and it reads 3 inches (It would read this if the hypotenuse was 21 inches as I calculated the proportion between the radius he found and the hypotenuse). Then we just divide our hypotenuse by the proportion (21/3) and we get exactly what he got: 5/7!

How do we know the hypotenuse is parallel to the joined radii of the circles?

ok.. this was kinda fun

I've spotted that because it's a similar triangle that you've constructed in the centre, you can immediately tell that it's sides are: 2r, 6/5*r, and 8/5*r. I think it's much easier this way. Although i think that your method is must more... methodical.

Using proportions x= 2r e y= 3r, so 7r=5.

I like watching this type of exercises even though I don't study maths

An alternate solution: It should be fairly trivial to demonstrate that the inradius of a 3-4-5 triangle is 1/5 of the hypoteneuse. Draw a triangle with lengths 3r, 4r, and 5r, draw its incircle, and drop the radii from the incenter to the tangent points. You can demonstrate that the radii form a square of length r, and kites with long edges of 2r and 3r. Take the diagram at 2:00 and note that the square and kites at the corners are similar to the square and corners from the triangle we just drew. Moreover, if the radius of the circles are set equal, the respective kites and squares are in fact congruent. The hypoteneuse of this triangle is thus 5r + 2r = 7r, which leads us rather directly to the result r = 5/7.

I considered that the center of the right-most circle would be part of the bisector of the bottom-right angle, and so i thought "if the center is Y away, horizontally, from the right vertex, then it must be (3/8)*Y tall (=> r=3Y/8)", which isn't consistent with your formulas. Why does my reasoning not make sense?

died when he said "and that's a good place too stop"

The general formula is r = R/(n+1)* 2c / (a+b)