# How to find the legs of a right triangle if its hypotenuse is numerically equal to its area?

I have an amazing ancient problem collection book by Chistyakov. I have managed to solve 165 problems of 248 (as of now, i. e. of time of current question posting); the remaining ones are really hard.

If you permit, I would like to ask help for problem #100.

One must find the legs (= catheti) of a right triangle, if it is known that its hypotenuse is numerically equal to its area.

So hypotenuse equals area (ignoring the units of measurement). Find the catheti!

[I suppose if there are multiple such triangles, then we must find all of them or find a general formula or something like that.]

How to solve such kind of geometrical problem?

• What is the name of the book? Can you tell me as well? Apr 19 at 7:59
• Let the cateti be $a$ and $b$, the hypotenuse $c$, and the area $A$. You are told $c = A$. Now, isn't there a formula for $A$ that connects it to $a$ and $b$? And isn't there a famous theorem that connects $a$, $b$ and $c$? If you find those, you got yourself a system of equations... Apr 19 at 8:02
• It's worth noting that since you're (numerically) equating an area with a length, scaling any right-angled triangle appropriately will give you a solution. If your initial triangle has hypotenuse $c$ and legs $a,b$, the ratio of the hypotenuse to the area is $\frac{2c}{ab}$. If the scaled triangle has corresponding sides $at,bt,ct$, the ratio becomes $\frac{2c}{abt}$; so choosing $t=\frac{2c}{ab}$ will give you a solution. Apr 19 at 12:24

Let $$ACB$$ be a triangle right angled at $$C$$ let $$AB=c, AC= b, BC = a$$. Then by Pythagoras theorem $$a^2+b^2=c^2$$ but given $$c=\frac{ab}{2}$$. Plugging in and simplifying you get $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{4}$$
Now using reciprocal Pythagoras theorem the lhs is equal to $$\frac{1}{d^2}$$ where $$d$$ is the altitude from the side $$AB$$ to the vertex $$C$$ solving you get $$d=2$$.
Now draw this altitude and using trigonometry find that $$b=\frac{2}{\cos (A)}$$ and $$a=\frac{2}{\sin (A)}$$

• Ahh yes thank you, I edited it now. Apr 19 at 8:22
• It should be $sin(α)$ and $cos(α)$, not $sin(A)$ and $cos(A)$. Apr 19 at 18:58

You have two equations:

$$c = \dfrac{1}{2} a b$$

and

$$c^2 = a^2 + b^2$$

So

$$\dfrac{1}{4} a^2 b^2 = a^2 + b^2$$

Let $$x = a^2, y = b^2$$ then

$$x y = 4 (x + y)$$

where $$x \gt 0 , y \gt 0$$

Hence,

$$x( y - 4) = 4 y$$

$$x = \dfrac{4 y}{y - 4} = 4 + \dfrac{ 16 }{y - 4}$$

For example, choose $$y = 5$$ , then

$$x = \dfrac{20}{5 - 4} = 20$$

Check:

$$b = \sqrt{5} , a = \sqrt{20} = 2 \sqrt{5}$$

$$c = \sqrt{ 5 + 20 } = 5$$

and $$Area = \dfrac{1}{2} ( 2 \sqrt{5} ) (\sqrt{5} ) = 5$$

So they're equal.

In the triangle $$ABC$$ right angled at $$C$$, call $$F$$ the foot of the perpendicular from the vertex $$C$$ on the hypotenuse. Denote the lengths of the mentionned segments $$c,a,b,h$$ respectivelly. The use of the "area=hypotenuse" constraint $$c=\frac{hc}{2}=\frac{ab}{2}$$ gives $$h=2\quad\text{and}\quad ab=2c.$$ There exist infinity of convenient triangles. The minimal area is $$4,$$ the triangle is then isosceles. There is no bound for the area maximum.

For the construction, I used Euclid altitude theorem, as it is called in my country. The corresponding identity $$|AF|\cdot |FB|=h^2$$ is due to similar right triangles $$AFC$$ and $$CFB.$$

The right angle of the triangle lies on a circle whose diameter is the hypotenuse.

In order for the area to equal the hypotenuse, the altitude on the hypotenuse is $$2$$.

These two facts exactly determine the shape of the triangle.

Let the length of the hypotenuse be $$c = AB$$ in the figure. Then $$OC = \frac c2$$. Let $$h = CM$$; then the area of triangle $$\triangle ABC$$ is $$\frac12 hc = c$$ and therefore $$h = 2$$.

By the Pythagorean Theorem on triangle $$\triangle CMO$$, then $$(MO)^2 = (OC)^2 - (CM)^2 = \frac{c^2}{4} - 4$$. Then \begin{align} (AC)^2 &= (AM)^2 + (CM)^2 \\ &= \left(\frac c2 - \sqrt{\frac{c^2}{4} - 4}\right)^2 + 4 \\ &= \frac{c^2}2 - c\sqrt{\frac{c^2}{4} - 4} \\ &= \frac c2\left(c - \sqrt{c^2 - 16}\right), \\ (BC)^2 &= (BM)^2 + (CM)^2 \\ &= \left(\frac c2 + \sqrt{\frac{c^2}{4} - 4}\right)^2 + 4 \\ &= \frac{c^2}2 + c\sqrt{\frac{c^2}{4} - 4} \\ &= \frac c2\left(c + \sqrt{c^2 - 16}\right). \\ \end{align}

[

TO FIND THE LEGS OF THE RIGHT TRIANGLE HAVING ITS HYPOTENUSE AS IT'S AREA

Consider ABC a triangle such that AC is the hypotenuse while AB (b) and BC (a) will be the sides (legs).

Now , $$AC = \frac{1}{2} ab$$

$$BC = AC \cosθ$$

$$a = 1 / 2 ab \cosθ$$

$$b = 2 / \cos θ$$

Similarly $$AB = AC \sinθ$$

$$b = 1 / 2 ab \sinθ$$

$$a = 2 / \sinθ$$

Therefore , $$b = 2 \secθ$$

$$a = 2 \operatorname{cosec}θ$$