Kvant Math Problem 556

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Verdicts: PASS + FAIL
Solve time: 31m46s
Source on kvant.digital

Problem

Must two acute isosceles triangles with equal lengths of the equal sides and equal radii of their inscribed circles necessarily be congruent?

A. A. Egorov

Solution

The answer is yes. The given data determine the triangle uniquely.

Let the equal sides of the acute isosceles triangle have length $a$, and let the vertex angle between them be $2\alpha$, where $0<\alpha<\frac{\pi}{4}$ because the triangle is acute.

The base then has length

$$b=2a\sin\alpha.$$

The area is

$$S=\frac12 a^2\sin 2\alpha =a^2\sin\alpha\cos\alpha.$$

The semiperimeter is

$$s=\frac{2a+b}{2} =a+a\sin\alpha =a(1+\sin\alpha).$$

Since the inradius $r$ satisfies $S=rs$, we obtain

$$r=\frac{S}{s} =\frac{a^2\sin\alpha\cos\alpha} {a(1+\sin\alpha)} =a\frac{\sin\alpha\cos\alpha}{1+\sin\alpha}.$$

Multiplying numerator and denominator by $1-\sin\alpha$ gives

$$r =a\frac{\sin\alpha(1-\sin^2\alpha)} {(1+\sin\alpha)(1-\sin\alpha)} =a\sin\alpha(1-\sin\alpha).$$

Hence

$$\frac r a=x(1-x), \qquad x=\sin\alpha.$$

Because the triangle is acute, $0<\alpha<\frac{\pi}{4}$, so

$$0<x<\frac{\sqrt2}{2}.$$

Consider the function

$$f(x)=x(1-x).$$

Its derivative is

$$f'(x)=1-2x.$$

For $0<x<\frac{\sqrt2}{2}$ we have $x<1/2$ or $x>1/2$. The critical point is $x=1/2$, and

$$f!\left(\frac{\sqrt2}{2}\right) =\frac{\sqrt2}{2}\left(1-\frac{\sqrt2}{2}\right) <\frac14 =f!\left(\frac12\right).$$

Thus every value attained by $f$ on the interval

$\left(0,\frac{\sqrt2}{2}\right)$ is attained exactly once. Indeed, if

$$f(x_1)=f(x_2),$$

then

$$x_1(1-x_1)=x_2(1-x_2),$$

which yields

$$(x_1-x_2)(1-x_1-x_2)=0.$$

Either $x_1=x_2$, or $x_1+x_2=1$. The second possibility cannot occur because both $x_1$ and $x_2$ lie in $\left(0,\frac{\sqrt2}{2}\right)$ and at most one of the two numbers solving $x(1-x)=c$ belongs to this interval. Hence $f$ is one-to-one on the admissible range.

Therefore the quantity $r/a$ uniquely determines $x=\sin\alpha$, and hence uniquely determines $\alpha$. Since $a$ is given and the vertex angle $2\alpha$ is uniquely determined, the triangle is uniquely determined up to congruence.

Consequently, two acute isosceles triangles with equal lengths of the equal sides and equal inradii must have the same vertex angle and the same side lengths. They are congruent. ∎