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

@sharkyigor

2 жыл бұрын

I just posted a comment with another pure algebra riddle.. if you want to give it a go.. :)

@pwnwin

2 жыл бұрын

Except at first you don’t know it’s just algebra and went crazy.

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.

@adityagarg3931

2 жыл бұрын

No the highest honour is if you "Sucon"

@michaeldatseris2149

2 жыл бұрын

That's not cheating on a maths test that's just learning your stuff (unless your teacher specified that you weren't allowed to do learning outside of the class which would be kind of weird )

@ValeriePallaoro

2 жыл бұрын

Is it a cheat if it's actually another way of working, even though it's more simple and concise? No, in the real world it's not a cheat (it's your teachers being *looks at Michael's comment from three days ago) Yeah, what he said.

@alexisbaliari

2 жыл бұрын

The whole point of maths in school is to train your brain to solve logical problems, and think. So it's not cheating.

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

@darealbeesechurger

2 жыл бұрын

Lol ikr

@lyrimetacurl0

2 жыл бұрын

And circles like Spongebob.

@_Mackan

2 жыл бұрын

I just wanna know how he does that

@TheXLAXLimpLungs

2 жыл бұрын

"And that's a good place to stop"

@tydshiin5783

2 жыл бұрын

@@_Mackan use your arm, never wrist

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.

@justanunverifieduser2075

2 жыл бұрын

pitots theorem ftw

@Ewr42

2 жыл бұрын

I """visualized""" the circle approaching the angled lines towards the direction of the single point of that angle and it seems to be the case that the center of the circle is always at the middle "height" point of the distance between the two lines. I can't expand to a full proof but I think there is one

@thehungrylittlenihilist

2 жыл бұрын

Thanks. I'm not very good at math. It seemed sort of intuitive but I couldnt tell you why or prove it.

@lyrimetacurl0

2 жыл бұрын

Oh, it just seemed obvious to me.

@adityachatterjee8870

2 жыл бұрын

@@thehungrylittlenihilist you’d use the same logic used in the video. Join the center of the circle to the external point. This gives you two triangles that are congruent, which proves the theorem

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

@nicholasong2892

2 жыл бұрын

I'm not japanese and i can confirm school math life was hell for everyone.

@kavishsingh1236

2 жыл бұрын

no culture or country ll survive without giving such hell to there future generations. This hell is the base they deserve it.

@ckwatt11

2 жыл бұрын

@@kavishsingh1236 this has to be the worst take I've ever seen. how dull of a person do you have to be to genuinely believe shit like this?

@stressedpanda7205

2 жыл бұрын

@@kavishsingh1236 Maths teaching doesn't have to be 'hell' any more. KZread is a great source of videos that clearly explain maths topics - and you can watch them as many times as it takes to grasp the concepts.

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

6:46

@somasahu1234

2 жыл бұрын

How much time u took to draw the circles ?

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

@wiktor3727

2 жыл бұрын

Yes, but the difficulty is like super high. If you see this puzzle first time it's nearly impossible (for most people I guess) to find solutions even if they know all rules that apply here (like why X and X are equal etc.) I highly recommend for you game for mobile phones called "euclidea", you may like it if you enjoy math

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

@dreadpiratewest6058

2 жыл бұрын

Glad I wasnt the only one!

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)]

@michellauzon4640

2 жыл бұрын

In general this method gives c = ( b / (c - a) + a / (c - b) + 2 ) * r Am I wrong or from this to your formula there is a long way ?

@nasekiller

2 жыл бұрын

@@michellauzon4640 i dont know what you did, but here is my way: We know that tan x/2 = sin x / (1+cos x) In one of the Angles that gives us tan = a/(b+c) in the other Tan= b/(a+c), since sin and cos are Just a/c and b/c respectively. Now this gives us x=(b+c)/a* r and y =(a+c)/b*r and this c=((a+c)/b+(b+c)/a+2)*r. The right Hand side can be rearranged to be (a+b+c)(a+b)/ab * r and then auch are basicallr done.

@michellauzon4640

2 жыл бұрын

@@nasekiller OK, nice! I used the other identity for tan(x/2). I didn't know the one you used.

@HoSza1

2 жыл бұрын

But c can be eliminated from that expression since it is a function of a and b.

@nasekiller

2 жыл бұрын

@@HoSza1 i mean, yeah, you could also eliminate a or b, but that doesnt really make the formula nicer.

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.

@yuriypr72

2 жыл бұрын

Exactly, this way there are no x and y involved. A much simpler solution.

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.

@pudgeboyardee32

2 жыл бұрын

Check out videos having to do with machining, machinists have to use geometry and trig very often and can offer tips and clues to simplify the math involved. The other upside beyond them working with those disciplines daily is that metal-working allows for the work to be directly shown and visually expressed. Its best when you can hold the work-piece for yourself but seeing the numbers being used on the page translate into an actual material can really help make geometry approachable. In the same vein, free computer-aided-drafting software for student use(AutoCAD is the example that springs to mind) allows for a virtual space to do all the same visualization and manipulation.

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

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

Excellent video!! Thank you!

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.

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.

@kefir321

2 жыл бұрын

Geometry actually hurts my head lol. Wish I was smart enough to understand math. Looks quite beautiful

So cool. Excellent explanation! Thanks.

that was really good, concise and easy to follow

Very elegant! I love it.

Your explanation is very clear.

Love these sangaku videos

You're a really great teacher

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.

thats really great

nice job, mike

Very satisfying and very well explained

Mindblowing!

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

Good one !

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?

@Ewr42

2 жыл бұрын

(2+11/5)r=3 => r=5/7?

@Glucenaphene

2 жыл бұрын

@@Ewr42 yes

@Ewr42

2 жыл бұрын

@@Glucenaphene I can't believe my stupidity, but maybe it's because they're All primes and I'm already bad enough at fractions to sum them up trivially Can you believe I actually went to college and had calculus twice?(not that I had to take it twice or that I didn't take any more calculus classes after that) which is completely believable

@Ewr42

2 жыл бұрын

Wow. Talk about NOT having impostor syndrome because you're actually worse than you think. I don't think I'll ever do any maths again after this

@Glucenaphene

2 жыл бұрын

@@Ewr42 calculator

This was really fun.

I was always wondering this

That was beautiful

This is so dope

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)

@henricobarbosa7634

2 жыл бұрын

I was thinking about that too. Or maybe it was just a nice coincidence

@cosimoraugei7325

2 жыл бұрын

@@henricobarbosa7634 I am solving it naming the three sides: a, b , sqrt(a^2+b^2)

@MMMM-br4pd

2 жыл бұрын

If the lengths of the catheti are a and b, the radius of each circle is (sqrt(a²+b²)-(a²+b²)/(a+b))/2

@sankalpsundar1668

2 жыл бұрын

no this is just a coincidence that works because 3+4 -5 =2, In genral if a,b are the sides and c is the hypotenuse, we have the radius of the circle is c(a+b-c)/(2(a+b))

@jpdiegidio

2 жыл бұрын

Nice exercise. Using sides a,b,c in place of 3,4,5 resp., I get r = [c*(a+b-c)] / [2*(a+b)]

Beautiful

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.

Cool shapes

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.

@vincentvapeur6559

Жыл бұрын

The most elegant solution !

Wow simply brilliant. 👏👍😊

Feels good to go back through geometric problem solving.

This channel feels like a massive flex

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

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

@l3p3

2 жыл бұрын

I give you one year with a maths book on an island and you would have figured that out.

@milanstevic8424

2 жыл бұрын

give me a million years and I wouldn't think of that part as hard or unfathomable, wtf

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.

Thank you..

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

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]

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

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

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

@vinegar6676

2 жыл бұрын

норм

@MM-vs2et

2 жыл бұрын

Well now you know! That’s really the first step to solving tricky puzzles like that. Use your imagination, within the boundaries of Euclidean geometry of course.

6:34 perfect

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

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!!

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)

Instant sub.

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

That is so beautiful gonna make me shed a teat

Goddamn, I had forgotten how much I loved math.

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.

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.

@delanmorstik7619

2 жыл бұрын

Well, your brain had few years to grow up and suddenly what seemed hard in the past is easy-peasy

Nice ✨✨

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

@jayc2483

2 жыл бұрын

From what I can tell it only works on a 3,4,5 style right-triangle. As you scale up the sides, you scale up r in the same proportion. Eg 2x to 6,8,10, r becomes 10/7 (which is 2x 5/7). Interestingly, hyp/(b+h) is constant at 5/7. None of the above was true for different proportioned of right triangles (eg a 5, 12,13). It's been so long since I've done any geometry or algebra though, so I might be doing it all wrong, apologies if so.

@tklalmighty

2 жыл бұрын

It may be true for some other triangles but definitely not all. Take a right triangle and try reducing one leg towards the absurd length of 0. Intuitively, the radius should also be reduced towards 0, but the ratio of the length of the hypoteneuse over the sum of the lengths of the legs trends to 1. Another simple counterexample outside of right triangles is an equilateral triangle. The radius of one of those circles is most definitely not half an edge length. Among right triangles, with identical circles inscribed and both tangent to the hypoteneuse, I'll assert that the radius of one of these circles is actually r = C / (A + B) × (A + B - C) / 2. The fact that the radius is equal to the length of the hypoteneuse over the sum of the lengths of the legs for the 3-4-5 triangle is coincidental because the sum of the lengths of the legs is 2 more than the length of the hypoteneuse.

@shortForPychael

2 жыл бұрын

@@tklalmighty that’s a good example with the limit as the base approaches 0, intuitive example that was easy to understand why it wouldn’t be the case for all triangles.

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.

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

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.

... 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.

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

@siddhantgarodia3381

2 жыл бұрын

@T143 hypotenuse not height

@siddhantgarodia3381

2 жыл бұрын

That's exactly what I was noticing, wondering if it works with other triangles too

@dylansp4049

2 жыл бұрын

@@siddhantgarodia3381 if that is the case, then the entire math problem is simple enough for a 1st grader to do.

@mmtasty1889

2 жыл бұрын

@@dylansp4049 the math problem is the proof rather than the calculation

@danmarzari7342

2 жыл бұрын

The general solution (in this problem A = 3, B = 4, C = 5) is R = (C^2 - AC - BC)/(-2 * (A+B)) So the (L1 + L2) part in the denominator was exactly your intuition, but the numerator gets complicated. Hope that helps!

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?

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

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

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).

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

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) )

@gbyt034

2 жыл бұрын

R= 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

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.

ok.. this was kinda fun

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

shapes in math. the bane of my existence

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.

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

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

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.

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)

the choice to not set the three equations up in an augmented matrix to solve for r made my spine itch.

I rate this video 5/7

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

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.

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

Holy crap

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?

@DonkoXI

2 жыл бұрын

It's a coincidence. There's another comment thread on this video where this is discussed in more detail.

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

My guess is 0.75 cause the circle looks like half the size of the smallest size of the triangle (½ x 3 = 1.5) and half of 1.5 is 0.75

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

@-danR

2 жыл бұрын

Huh. Ice-cream hadn't even been _invented_ in Euclid's day. Cool 🍨 story, dude.

@Jack_Callcott_AU

9 ай бұрын

​@@-danR But in Euclid's day they might have had goat cream.

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

Nice explanation, thanks. Is it possible to write r=H/A+O as a rule? No, just thought it through with smaller circles, thanks.

Wow

Tell me why ain´t nothin` but a heartache