It's a high school level question which we can't seem to solve. Here it is:

Given $2$ lines, one of the length of the hypotenuse and the other with the length of the sum of the $2$ legs, construct with straightedge and compass the corresponding right triangle

I didn't make much progress. It seems that there's a theorem or a few basic facts about right triangles that I'm missing. What path do you suggest I take?

Thanks for your help.

  • $\begingroup$ Does this mean that you are given the length of the hypotenuse, and the sum of the lengths of the two legs, and you are supposed to find a triangle that satisfies those constraints? When it says "construct", does it mean "construct with straightedge and compass", or what? $\endgroup$ – MJD Oct 8 '12 at 17:06
  • $\begingroup$ yes, I've edited my question - I hope it's more clear now. $\endgroup$ – Amihai Zivan Oct 8 '12 at 17:12

Let $ABC$ be the triangle you want to construct, with $\angle A =90^\circ$.

Then you are given $AB+AC$ and $BC$.

Extend $BA$ past $A$ to $BB'$ by a length equal to $AC$, that is $AB'=AC$. Then, the triangle $ACB'$ is a right isosceles triangle.

This means that in the triangle $BB'C$ you know $BB'=AB+AC$, $BC$ and the angle $B'=45^\circ$.

This suggests how you can construct it: construct the triangle $BB'C$, and then construct the height from $C$. The leg of the height will be $A$.

Since you are constructing $BB'C$ by $SSA$, there should be two solutions for $C$.

Here is the actual construction:

Start by drawing an angle of $45^\circ$. Denote the vertex of the angle $B'$.

On one side pick a point $B$ so that $BB'=AC+AB$.

Next draw a circle of centre $B$ and radius $BC$. This will intersect the other ray of the angle in two points $C_1, C_2$. Pick the one which makes the angle $CBB'$ acute.

P.S. I always solve the problems the way i.m.s. did, I like more the algebraic approach, but since you mention that this is high school level you are probably looking for the geometric approach.


You know the hypotenuse length $c$, and the sum of the lengths of the legs $u=a+b$. Thus you also know $v=u^2-c^2=a^2+b^2+2ab-a^2-b^2=2ab$. Then $b=u-a=u-v/2b$ so b satisfies the quadratic equation $b^2-ub+v/2=0$ which can be constructed using ruler and compass.


The algebra can guide the geometry. We want to construct a triangle with given hypotenuse $c$ and legs $a$ and $b$, where we are given $a+b$.

Note that $$(a-b)^2=2(a^2+b^2)-(a+b)^2=2c^2-(a+b)^2.$$ It is easy to construct a line of length $\sqrt{2} c$. Then construct the right triangle with hypotenuse $\sqrt{2}c$ and one leg equal to $a+b$. Easy, with center the middle of the hypotenuse, draw a circle that passes through the endpoints. Then through one endpoint draw a circle of radius $a+b$.

The other leg of our just constructed triangle has length $|a-b|$. Subtract this from $a+b$. The result is twice a leg of our target triangle.


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