An example of classic Japanese geometry -- Sangaku
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- เผยแพร่เมื่อ 5 ต.ค. 2024
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Nice to have a little puzzle like this that's just algebra once in a while. :)
I just posted a comment with another pure algebra riddle.. if you want to give it a go.. :)
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.
No the highest honour is if you "Sucon"
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 )
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.
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
Lol ikr
And circles like Spongebob.
I just wanna know how he does that
"And that's a good place to stop"
@@_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.
pitots theorem ftw
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
Thanks. I'm not very good at math. It seemed sort of intuitive but I couldnt tell you why or prove it.
Oh, it just seemed obvious to me.
@@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.
I'm not japanese and i can confirm school math life was hell for everyone.
no culture or country ll survive without giving such hell to there future generations. This hell is the base they deserve it.
@@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?
@@kavishsingh1236 Maths teaching doesn't have to be 'hell' any more. TH-cam 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
I always loved constructing and solving for variables like this in highschool geometry. It's basically just puzzle solving with math!
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.
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)]
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 ?
@@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.
@@nasekiller OK, nice! I used the other identity for tan(x/2). I didn't know the one you used.
But c can be eliminated from that expression since it is a function of a and b.
@@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.
Exactly, this way there are no x and y involved. A much simpler solution.
6:46
How much time u took to draw the circles ?
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.
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 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.
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.
The most elegant solution !
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.
Geometry actually hurts my head lol. Wish I was smart enough to understand math. Looks quite 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?
(2+11/5)r=3 => r=5/7?
@@Ewr42 yes
@@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
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
@@Ewr42 calculator
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)
I was thinking about that too. Or maybe it was just a nice coincidence
@@henricobarbosa7634 I am solving it naming the three sides: a, b , sqrt(a^2+b^2)
If the lengths of the catheti are a and b, the radius of each circle is (sqrt(a²+b²)-(a²+b²)/(a+b))/2
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))
Nice exercise. Using sides a,b,c in place of 3,4,5 resp., I get r = [c*(a+b-c)] / [2*(a+b)]
I'm not a math freak at all, but this is undoubtedly beautiful.
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.
This channel feels like a massive flex
Thank you for these videos. I look at one each day at work. Very interesting and brain excercising.
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
норм
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.
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]
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)
Excellent video!! Thank you!
You're a really great teacher
Your explanation is very clear.
This is a way better way of demonstrating system of equation technique than anything we learned in algebra class, thank you
"And that's a good place to stop" this guy can write a novel
Cool how it ends up being hypotenuse/(base+height)
I wonder if that’s true for other triangles?
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.
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.
@@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.
Interesting that the radius is H/(L1+L2)
@T143 hypotenuse not height
That's exactly what I was noticing, wondering if it works with other triangles too
@@siddhantgarodia3381 if that is the case, then the entire math problem is simple enough for a 1st grader to do.
@@dylansp4049 the math problem is the proof rather than the calculation
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!
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.
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) )
R= c/a+b
Seems the general case for a right triangle with legs a and b (a
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)
For general a,b c we have r = a*(b+c-a)/2*(b+c)
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.
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.
So cool. Excellent explanation! Thanks.
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.
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.
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
Is it obvious that the rectangle drawn is indeed a rectangle (i.e. that the segment joining the centres of the circles is parallel to the hypothenuse of the big triangle)? This is a crucial step and it looks to me that this fact is just given, not proven.
it is parallel because the sides of the rectangle are "r" long, thus being equal length.
and the fact that the hypotenuse is perpendicular to the radius can be proved by the property [radius (is) perpendicular (to) tangent].
Since there is a pair of opposite sides with equal length (the radii) and that the adjacent angles are 90 degrees, the quadrilateral must be rectangle.
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
That's a great puzzle, I will rate it 5/7
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.
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.
Goddamn, I had forgotten how much I loved math.
This was really fun.
Very elegant! I love it.
Very satisfying and very well explained
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
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
The answer to (1) is lower down the thread. There are two equivalent solutions:
r = c/(a+b) * (a+b-c)/2
or
r = abc / [(a+b)(a+b+c)]
Love these sangaku videos
If you see closely fro any perpendicular triangle u can divide the hypotenuse with the sum of the adjacent sides
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 was always wondering this
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.
The radius is: draw it in solidworks and then use the measure tool
Basically where im at these days. I understood all the geometry he did, but my algebra and even basic arithmetic is so dismal these days i still couldnt follow towards the end
@@chriso1373 but at the same time, it's like walking when you can drive. Why spend all this time doing all this geometry, when you can just summon the answer in seconds.
@@t_c5266 definately the more convenient means to the end... still wish i hadnt lost all that math tho. Never know when you might need it 😥
thats really great
6:34 perfect
Feels good to go back through geometric problem solving.
For a 3:4:5 ratio triangle...
R = Hypoteuse / ( Height + Base)
R = 5 /( 3 + 4 ) = 5/7
nice job, mike
Mindblowing!
Wow simply brilliant. 👏👍😊
shapes in math. the bane of my existence
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?
It's a coincidence. There's another comment thread on this video where this is discussed in more detail.
Beautiful
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.
Well, your brain had few years to grow up and suddenly what seemed hard in the past is easy-peasy
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.
This is so dope
Cool shapes
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
have no idea why this was recommended but i enjoyed it, thank you
that was really good, concise and easy to follow
Thank you..
Instant sub.
The general formula is
r = R/(n+1)* 2c / (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?
died when he said "and that's a good place too stop"
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?
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.
I'm mathematically challenged, how you figured that out was amazing
1:39 wait, how can we be sure that this is indeed a straight line of 2r range? What if it angles at the middle point?
the circles are tangent to each other
@@BigDBrian 🤦♂️ right, of course
Yup, i am thinking about it too. U can try remove the triangle, just draw the two equal circles and the line, and u will see it
That's what I was wondering too... I would've said, measure ankles to find that out but then you can just measure the radius and the sides, so measuring shouldn't be allowed
I rate this video 5/7
Good one !
The theorem Michael quoted about tangents to a circle from outside the circle re x and y, is called "the ice-cream cone theorem"
Huh. Ice-cream hadn't even been _invented_ in Euclid's day.
Cool 🍨 story, dude.
@@-danR But in Euclid's day they might have had goat cream.
That is so beautiful gonna make me shed a teat
That was beautiful
the choice to not set the three equations up in an augmented matrix to solve for r made my spine itch.
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
I got a tape measure if you need to burrow it might be a little quicker
Good explanation, just clicked for fun and stay 😂
Using proportions x= 2r e y= 3r, so 7r=5.
35 years ago I sat in school and a temp came up to me ( interested in math ) and gave me a mathematical problem. I have never solved it. Those I given it have failed too. It is similar to the above problem. And it seems so easy. "You have a right triangle with hypotenuse of length 10. There is a square (3*3 units ) inside the triangle, two sides are along the legs and the opposing corner is touching the hypotenuse. How long are the legs!" Impossible to solve, apparently. Would you like to try it out? ( I guess there is some simple solution that I never been able to see.)
I think it could be proven that the problem is impossible and that would be a satisfying end. You can draw that accurately on paper using many different length of legs, which I believe means that the problem is indeed impossible, but it is not valid proof for it.
@@arttutaipale7696 If you try drawing this problem on a paper (or draw it in Geogebra), you'll find that the legs have to be approximately 8.9 and 4.5 in length.
By drawing it you now have three right triangles of equal form/ratios to work with. The sides of the largest triangles can be written as (3+x)^2 + (3+y)^2 = 10^2.
x or y can be substituted so that you only have one unknown to work with, but you'll end up with a long and complicated formula with x^4 etc. Not sure where to go from here. 😐
I solved somewhat like in video. But harder ))
The beauty is here: 5 / 7 = 5 / (3 + 4)
How can you prove that the longest side of the 4-corner figure (looks like a rectangle) is equal to 2r? It's need to prove that this figure is really a rectangle for this purpose.
Tell me why
ain´t nothin` but a heartache