# Problem about right triangles.

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.

• 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? – Gerry Myerson Oct 12 '14 at 6:47
• @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? – Jackie Poehler Oct 12 '14 at 7:20
• Here's another possible way forward: start by proving $a^2+b^2=c^2$ implies $a+b<c\sqrt2$. – Gerry Myerson Oct 12 '14 at 7:51
• @GerryMyerson stuck on proving it. How do you do it? assume angles are 45 45 and 90? – Jackie Poehler Oct 12 '14 at 14:37
• no it's all clear and beautiful . thanks – Jackie Poehler Oct 15 '14 at 13:34

## 3 Answers

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:

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 ($c_i$).)

• Please further explain the past paragraph. I am finding it difficult to understand. – anonymous Feb 25 '15 at 7:00

I can only give the following suggestion and its partial proof.

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.

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.

• What is v equal to? What is the triangle inequality? – anonymous Feb 25 '15 at 6:57
• @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. – Gerry Myerson Feb 25 '15 at 21:57