Two right-angled triangles

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Two right-angled triangles

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Source: French TST 2005 - pb 2

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Igor

#1 h May 26, 2005, 5:04 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one. Let $S$ (respectively $S'$ ) be the area of the first triangle (respectively of the second triangle).

Prove that $\frac{S}{S'}\geq 3+2\sqrt{2}$ .

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Igor

#2 h May 26, 2005, 7:53 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

My solution:
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#3 h May 26, 2005, 11:25 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Igor wrote:

Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one.
Let $S$ (respectively $S'$ ) be the area of the first triangle (respectively of the second triangle).
Prove that $\displaystyle \frac{S}{S'}\geq 3+2\sqrt{2}$ .

It seems to be difficult to give a non-bash solution of this problem, but let me try:

Let ABC be the first right-angled triangle, with the right angle at C, and with the sides a = BC, b = CA, c = AB and the angles A = < CAB, B = < ABC, C = < BCA. Let A'B'C' be the second right-angled triangle, with the right angle at C', and with the sides a' = B'C', b' = C'A', c' = A'B' and the angles A' = < C'A'B', B' = < A'B'C', C' = < B'C'A'. According to the conditions of the problem, the inradius r of triangle ABC equals to the circumradius of triangle A'B'C'.

The circumradius of a right-angled triangle equals $\frac12$ of its hypotenuse. Since the triangle A'B'C' is right-angled with the hypotenuse c', its circumradius thus equals $\frac{c^{\prime}}{2}$ ; on the other hand, we know that its circumradius is r. Thus, $r=\frac{c^{\prime}}{2}$ , so that 2r = c'. But by the Pythagorean theorem, applied to the right-angled triangle A'B'C', we have $c^{\prime}\ ^2=a^{\prime}\ ^2+b^{\prime}\ ^2$ . On the other hand, the area S' of the right-angled triangle A'B'C' equals $\frac12$ of the product of its catets a' and b'; in other words, $S^{\prime}=\frac12 a^{\prime}b^{\prime}$ . By the inequality $2xy\leq x^2+y^2$ for any two real numbers x and y, we thus have

$4S^{\prime}=4\cdot\frac12 a^{\prime}b^{\prime}=2a^{\prime}b^{\prime}\leq a^{\prime}\ ^2+b^{\prime}\ ^2=c^{\prime}\ ^2=\left(2r\right)^2=4r^2$ ,

so that $S^{\prime}\leq r^2$ .

On the other hand, by Lemma 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 post #19, we have

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$ ,

since S is the area and r is the inradius of triangle ABC. Since C is a right angle, we have $\frac{C}{2}=\frac{90^{\circ}}{2}=45^{\circ}$ , so that $\cot\frac{C}{2}=\cot 45^{\circ}=1$ , and this equation becomes

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}$ .

The angles A and B are the two acute angles of the right-angled triangle ABC; thus, they satisfy 0° < A < 90° and 0° < B < 90°. Consequently, $0^{\circ}<\frac{A}{2}<45^{\circ}$ and $0^{\circ}<\frac{B}{2}<45^{\circ}$ . Consequently, by the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=21384 , we have

$\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}$ .

Since the triangle ABC is right-angled with the right angle at C, we have A + B = 90°, so that $\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}=\tan\frac{A+B}{4}=\tan\frac{90^{\circ}}{4}=\tan 22,5^{\circ}=\sqrt2-1$ , and thus this becomes

$\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq \sqrt2-1$ .

Since both sides of this inequality are nonnegative, we can square it:

$\tan\frac{A}{2}\tan\frac{B}{2}\leq\left(\sqrt2-1\right)^2$ .

Hence,

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}=\frac{r^2}{\tan\frac{A}{2}\tan\frac{B}{2}}\geq \frac{r^2}{\left(\sqrt2-1\right)^2}=r^2\cdot\left(3+2\sqrt2\right)$ .

Combining this with the inequality $S^{\prime}\leq r^2$ obtained before, we get

$\frac{S}{S^{\prime}}\geq\frac{r^2\cdot\left(3+2\sqrt2\right)}{r^2}=3+2\sqrt2$ ,

and the problem is solved.

Darij

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Sailor

#4 h May 28, 2005, 1:16 PM • 1 Y

Y by Adventure10

This one seems a bit easier:

Let c, and c' be the hypotenuses of the two triangles.
We have $a+b=c+c'$ .

We shall prove that:
$p\geq{h'}\cdot(3+2\sqrt{2})$ or
$p^2\geq{{h'}^2}(17+12\sqrt{2})$ (1).
Since ${h'}^2\leq{{c'}^2}\frac{1}{4}$ (1) becomes:
$(2c+c')^2\geq{(c')^2(17+12\sqrt{2})}$ . This reduces(after opening the paranthesis and factorising) to:
$c\geq{c'(1+\sqrt{2})}$ which is true because if not then:
$a+b=c+c'>c\cdot{\sqrt{2}}$ or $2ab>c^2=a^2+b^2$ contradiction!

Hence the inequality is proved!

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#5 h Jan 5, 2020, 6:54 PM • 1 Y

Y by Adventure10

Let $R'=r=t$ , hence $c'=2t$ and $4t^2=a'^2+b'^2\geq 2a'b'=4S' \implies S'\leq t^2$
$ab=2S=2rs=t(a+b+\sqrt{a^2+b^2})\geq t(2\sqrt{ab}+\sqrt{2ab})$ , we used AM-GM and AM-RMS.
So $\sqrt{ab}\geq t(2+\sqrt{2}) \implies S\geq t^2(1+\sqrt2)^2$
Combining both we have $\displaystyle\frac{S}{S'}\geq (1+\sqrt2)^2=3+2\sqrt{2}$
Equality iff both triangles are isoceles.

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