Let the given hypotenuse be the diameter $AB$ of the semicircle $O$. Every triangle $ACB$ inscribed in this semi-circle has a right angle at $C$ and a median $CO$ of length $\dfrac{c}{2}$. Our task is to locate the point $C$ in such a way that $CO$ is the geometric mean of $a=CB$ and $b=AC$, i.e. \[ (1) \Leftrightarrow \dfrac{a}{\dfrac{c}{2}}=\dfrac{\dfrac{c}{2}}{b} \] or \[ 4ab=c^2 \] We accomplish this by determining the altitude $h$ from the vertex $C$ of the desired triangle. On the one hand, the area of this triangle is $\dfrac{ab}{2}$ and also $\dfrac{ch}{2}$; hence $ab=ch$. On the other hand, equation $(1)$ must hold. Thus \[ ch=ab=\dfrac{c^2}{4} \Leftrightarrow h=\dfrac{c}{4} \] We construct a line parallel to the diameter $AB$ and at a distance $\dfrac{c}{4}$ above it. This line intersects the semicircle in two points $C$ and $D$, which are the vertices of the right triangles satisfying the requirements of the problem.

My proof is quite different. I approached it using the median of the traingle. Here it goes: Let $a$ and $b$ be the legs of the traingle. $a^2+b^2=c^2$ Using cosine law, we get, $b^2=ab+\frac{c^2}{4}-c\sqrt{ab}\cos\theta$ $\\a^2=ab+\frac{c^2}{4}-c\sqrt{ab}\cos (180^{/circ}-\theta)\\ a^2=ab+\frac{c^2}{4}+c\sqrt{ab}\cos\theta$, where $\theta$ is the angle between the median and the top half of the hypotenuse. $\\a^2+b^2=c^2\\ 2(ab+\frac{c^2}{4})=c^2\\ 4ab=c^2\\ c=2\sqrt{ab}$ So now, we have two iscoceles traingles creating a right traingle with the hypotenuse being $2\sqrt{ab}$, the median being $\sqrt{ab}$ with legs $a$ and $b$. Masoud Zargar

Sorry to revive an ancient topic, but I was doing this problem today, and saw that I have a different solution.

i also have a bit diff soln draw a line segment of given length "c". find the mid point(can be easily done using compass by making a perpendicular bisector). from the mid point take a arc of radius ${{c/2}}$ draw a circle now we have a circle with a diameter of length c now if we draw any angle in the semicircle we wud get a right triangle. so now we use ${a^2+b^2=4ab)}$ and denote ${a/b}$ by tanX for some X now we find that tanX=${2+sqrt3 or 2-sqrt3}$ now intuitively we find tan(30+45)=tanX hence one of the angles of the triangle is 75 degree which can be easily constructed from one end of the diameter.. hence if we join the point where this angle intersects the circle to the other end of the diameter.....we get our required triangle

I have only one example. It is with legs $ 2 - \sqrt{3} $ and $ 1 $. I can't find them using a and b. I am too young.

@DPopov: The right-angled triangle having the altitude a quarter of hypotenuse has an angle of $15^\circ$. For a proof, see the attached picture. Best regards, sunken rock

Can someone explain how does sunken rock's picture work? I can't understand it.

I am really sorry for the bump but would it be sufficient to construct a 45-45-90 triangle?

I dunno the answer but i just want to state that this topic started in 2005

I do not understand the problem and have a question. I found a contradiction. In the image shown below, $\bigtriangleup{ADC}$ and $\bigtriangleup{BDC}$ are congruent isosceles right triangles, and therefore, $AB=BC=\sqrt{2}BC$. Therefore, $BC^2=2BC^2$, and this is a contradiction. Can someone explain this?

The problem asks to construct one such triangle, not show that it is true for all right triangles.

Let the legs of the triangle be $a$ and $b$, with $a$ opposite from vertex $A$, $b$ opposite from vertex $B$, and $c$ opposite from the hypotenuse. We have that, since the length of the median to the hypotenuse is $\frac{c}{2}$, $$\frac{c}{2} = \sqrt{ab} \implies c^2=4ab \implies a^2+b^2=4ab \implies b = \frac{4a \pm 2\sqrt{3}a}{2} \implies b = \left(2+\sqrt{3} \right)a$$Therefore, $\tan CBA = 2+\sqrt{3}=\tan 75$ and $\angle CBA = 75 \implies \angle CAB = 15$. It’s hard to explain construction, but we can construct a $15$ degree angle by bisecting a $60$ degree angles twice. We can draw the $60$ degree angle using equilateral triangles.

We will show that only a $90^\circ-15^\circ-75^\circ$ triangle works due the length condition. Call the triangle $\triangle ABC$ and let it be right angled at $B$. Let $y$ be the circumradius of $\triangle ABC$. It is well known that $BM=AM=MC$. The condition states that $BM^2=BA \cdot BC$. Let $\angle BAC=\theta$. By definition, $BC=2y\sin(\theta)$ and $BA=2y\cos(\theta)$. Putting this in into the length condition, we get \[y^2=4y^2\sin(\theta)\cos(\theta)\]Cancelling $y^2$ off, and using $\sin(2\theta)=2\sin(\theta)\cos(\theta)$, we get \[\sin(2\theta)=\frac{1}{2}\]this goes down to $2 \theta = 30^\circ, 150^\circ \implies \theta = 15^\circ$ or $75^\circ$. One can assume without loss of generality that $\theta =15^\circ$. The construction of this angle is fairly well-known, so we are done. $\blacksquare$

Lol this thread is older than me

I have discussed this problem on my YouTube channel. Link: Video Solution

If $\theta$ is the least angle of triangle we have $\sin (2\theta )$ $=2\sin (\theta) \cos (\theta)=\frac{2c^2}{4c^2}=\frac{1}{2}$ $\implies$ $\theta =15^\circ.$ Now we can construct triangle by lenght of side and all angles (clearly we can construct angle with measure $15^\circ$).

Let $a$ and $b$ be the leg lengths. Then, the median to the hypotenuse has length $\tfrac{c}{2}$, while the geometric mean of the legs is $\sqrt{ab}$, so $\tfrac{c}{2}=\sqrt{ab}\iff\tfrac{c^2}{4}=ab\iff\tfrac{c^2}{8}=\tfrac{ab}{2}$ (because $a,b,c>0$). But since $\tfrac{ab}{2}$ is the area of the triangle, this is equivalent to the altitude to the hypotenuse having length $\tfrac{c}{4}$. Hence, begin with two points $A$ and $B$ that are $c$ units apart, and let $O$ and $\omega$ be the midpoint of $\overline{AB}$ and a semicircle with diameter $AB$ (either semicircle suffices). Further, let $P$ be the midpoint of arc $AB$ and $Q$ the midpoint of $\overline{OP}$. Letting $C$ be an intersection of $\omega$ and the perpendicular bisector of $\overline{OQ}$ (either intersection suffices) yields the desired triangle in triangle $ABC$, since $AB=c$, $\angle ACB=90^\circ$ (since $C$ lies on $\omega$), and $\text{dist}(C,\overline{AB})=\text{dist}(R,\overline{AB})=\tfrac{c}{4}$ (where $R$ is the midpoint of $\overline{OQ}$). $\blacksquare$

franchester wrote: I am really sorry for the bump but would it be sufficient to construct a 45-45-90 triangle?

)