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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?

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    $\begingroup$ What is the name of the book? Can you tell me as well? $\endgroup$ Apr 19 at 7:59
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    $\begingroup$ 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... $\endgroup$ Apr 19 at 8:02
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    $\begingroup$ 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. $\endgroup$ Apr 19 at 12:24

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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 $a=\frac{2}{\sin (A)}$

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

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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 $CFB.$ enter image description here

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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.

enter image description here

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}

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[I'll use this triangle reference2

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 $

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