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Given N>1 right triangles. Sum one legs of each of them, then sum all the left legs, then sum all hypotenuses. These 3 sums form the sides of a right triangle. Prove all given N triangles are similar

I'm really stuck on this one. Got that sum of all hypotenuses is the end hypotenuse. but can't proceed from it. Any help would be appreciated.

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  • $\begingroup$ Can you prove that if $a^2+b^2=c^2$ and $d^2+e^2=f^2$ then $(a+d)^2+(b+e)^2\le(c+f)^2$ with equality only in the similar triangles case? $\endgroup$ – Gerry Myerson Oct 12 '14 at 6:47
  • $\begingroup$ @GerryMyerson Well I ve gotten that if they are similar then cf=ad+be but I really cant prove that otherwise cf would not be equal to ad+be. How do you do that? $\endgroup$ – Jackie Poehler Oct 12 '14 at 7:20
  • $\begingroup$ Here's another possible way forward: start by proving $a^2+b^2=c^2$ implies $a+b<c\sqrt2$. $\endgroup$ – Gerry Myerson Oct 12 '14 at 7:51
  • $\begingroup$ @GerryMyerson stuck on proving it. How do you do it? assume angles are 45 45 and 90? $\endgroup$ – Jackie Poehler Oct 12 '14 at 14:37
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    $\begingroup$ no it's all clear and beautiful . thanks $\endgroup$ – Jackie Poehler Oct 15 '14 at 13:34
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Consider the contrapositive statement:

If the right triangles aren't similar, then their edge-sums don't form a right triangle.

Let triangle $i$ have legs $a_i$ and $b_i$, and hypotenuse $c_i$. We may assume that our triangles are ordered such that $$\frac{b_i}{a_i} \;\leq\; \frac{b_{i+1}}{a_{i+1}}$$

That is, laying them out as in the figure, the hypotenuses have steeper and steeper "slopes" as $i$ increases:

enter image description here

The combined legs form legs of a large right triangle with hypotenuse $d$. The chain of hypotenuses form a polygonal path from one endpoint of the large hypotenuse to the other; clearly(?), then $$d \;\leq\; c_1 + c_2 + c_3$$ with equality if and only if each $c_1$ lies on $d$. (We can make this more formal by dropping perpendiculars from the endpoints of the $c_i$ onto $d$, sub-dividing the large hypotenuse into segments of length $d_i$, where clearly(!) $d_i \leq c_i$ by the fact that a leg (here, $d_i$) of a right triangle is no longer than the hypotenuse ($c_i$).)

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  • $\begingroup$ Please further explain the past paragraph. I am finding it difficult to understand. $\endgroup$ – anonymous Feb 25 '15 at 7:00
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I can only give the following suggestion and its partial proof.

enter image description here

We will prove the claim by induction.

Let consider the case when $N = 2$. Also, we will restrict all angles to be less than or equal to $90^0$.

Then, according to the given hypothesis, we have $(a_1 + a_2)^2 + (b_1 + b_2)^2 = a^2 + b^2 = c^2$

After expansion and cancellation, we have $\frac {a_1a_2}{c_1c_2} + \frac {b_1b_2}{c_1c_2} = 1$

i. e. we have $\frac {a_1}{c_1}\frac {a_2}{c_2} + \frac {b_1}{c_1}\frac {b_2}{c_2} = 1$

i.e. we have $(\cos B_1)(\cos B_2) + (\sin B_1)(\sin B_2) = 1$

For the above to be true, B has to equal to B’ according to the identity ($\sin^2 x + \cos^2 x = 1$).

Thus, $⊿A_1B_1C_1$ ~ $⊿A_2B_2C_2$.

Now assume that it is true for $N = k$

We are now left with 3 triangles (because all others are similar already), namely $⊿A_{k+1}B_{k+1}C_{k+1}$, $⊿A_{S(k)}B_{S(k)}C_{S(k)}$, and $⊿A’B’C’$ with the following explanation of the notation used.

The side opposing $A_{S(k)}$ has length = sum of all $a_i$ for $i = 1, 2, 3, .... k$. $B_{S(k)}$ and $C_{S(k)}$ should be interpreted similarly.

The side opposing A' has length = sum of all $a_i$ for $i = 1, 2, 3, .... k, k+1$. Therefore, it length is equal to $S(k) + a_{k+1}$. B' and C' should be interpreted similarly.

The argument should be (and I hope) no different than the above in completing the induction proof.

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The solution by Blue is very nice. Here's much the same proof, but with equations instead of the pretty picture.

The triangle inequality says $$|v_1+\cdots+v_n|\le|v_1|+\cdots+|v_n|$$ with equality if and only if the $v_i$ are all scalar multiples of each other. Now let $v_i=(a_i,b_i)$, and let $a_i^2+b_i^2=c_i^2$, so $|v_i|=c_i$. Then $$v_1+\cdots+v_n=(a_1+\cdots+a_n,b_1+\cdots+b_n)$$ and the triangle inequality becomes $$\sqrt{(a_1+\cdots+a_n)^2+(b_1+\cdots+b_n)^2}\le c_1+\cdots+c_n$$ Squaring both sides, we have $$(a_1+\cdots+a_n)^2+(b_1+\cdots+b_n)^2\le(c_1+\cdots+c_n)^2$$ with equality if and only if all the $(a_i,b_i)$ are scalar multiples of each other, which is to say, if and only if all the $(a_i,b_i,c_i)$ right triangles are similar, and we're done.

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  • $\begingroup$ What is v equal to? What is the triangle inequality? $\endgroup$ – anonymous Feb 25 '15 at 6:57
  • $\begingroup$ @anon, I don't know what $v$ is --- I never used $v$ anywhere in my answer. I did use $v_1,\dots,v_n$, and those are all vectors; in the application, they are all ordered pairs of numbers, $v_i=(a_i,b_i)$, where the $a_i$ and $b_i$ are the legs of right triangles. The triangle inequality says a side of a triangle is no longer than the other two sides added together; the algebraic formulation of this geometric fact is $|v+w|\le|v|+|w|$. But, you know, you probably would have gotten the answer faster by just typing "triangle inequality" into a search engine. $\endgroup$ – Gerry Myerson Feb 25 '15 at 21:57

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