Geometric Inequality with Excircle

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Geometric Inequality with Excircle

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Source: Turkey National Mathematical Olympiad 2018

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#1 h Dec 2, 2018, 8:39 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , the bisector of the angle $A$ intersects the excircle that is tangential to side $[BC]$ at two points $D$ and $E$ such that $D\in [AE]$ . Prove that,
$\frac{|AD|}{|AE|}\leq \frac{|BC|^2}{|DE|^2}.$

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#2 h Dec 2, 2018, 9:09 PM • 2 Y

Y by Adventure10, Mango247

Here is a trigonometric solution.

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XbenX

#3 h Dec 3, 2018, 8:59 AM • 4 Y

Y by teomihai, Adventure10, Mango247, Mango247

Let $B', C'$ be points on lines $AB$ and $AC$ respectively, such that $B'C' \parallel BC$ and $B'C'$ is also tangent to $A$ excircle.
It is clear that there is a homothethy centered at A sending $\triangle ABC$ to $\triangle AB'C'$ and from the fact that $\frac{r_a}{r}=\frac{s}{s-a}$ we have that this homothethy has a coefficient of $\frac{s}{s-a}$ (here $a=\overline{BC}$ and $s$ semiperimeter).

Let the intersection of the line $AB$ and $A$ -excircle be $X$ . It is a well-known fact that $AX=s$ .

Now, by the power of point $A$ w.r.t to $A$ -excircle we get that $AD \cdot AE=AX^2 \Longleftrightarrow \frac{AD}{AE}= \frac{AX^2}{AE^2}$ .

Lemma: In every triangle $\triangle ABC$ let $I$ be the incenter and let the angle bisector at $A$ intersect the incircle at 2 points $P, Q$ such that $AQ>AP$ then $AQ>h$ where $h$ is the altitude erected from $A$ .

Proof of lemma: Let $Y$ be the foot of $I$ into $BC$ , we have that $IY=IQ$ and by triangle inequality $AQ=AI+IQ=AI+IY \geq AY \geq h$ .

Now by the above lemma in triangle $AB'C'$ we have that $AE\geq h$ (here $h$ is now altitude from $A$ to $B'C'$ ).

The area of $\triangle AB'C'$ is $r_a \cdot \frac{s^2}{s-a}= \frac{A'B' \cdot h }{2}= \frac {AB \cdot s \cdot h}{2(s-a)} \leq \frac{AB \cdot AE \cdot s}{2(s-a)}$ This gives that $2r_a \cdot s \leq BC \cdot AE$ and since $2r_a=DE$ and $s=AX$ this is equivalent to $\frac{AX}{AE} \leq \frac{BC}{DE}$ and we have that $\boxed{\frac{AD}{AE}=\frac{AX^2}{AE^2} \leq \frac{BC^2}{DE^2}}$ $\blacksquare$

Equality at $AE=h$ (i.e when $AB=AC$ )

This post has been edited 5 times. Last edited by XbenX, Sep 17, 2019, 12:24 PM

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#4 h Dec 29, 2021, 6:24 PM • 1 Y

Y by teomihai

Claim: If $|AB|=|AC|$ , then $\frac{|AD|}{|AE|}=\frac{|BC|^2}{|DE|^2}$ .
Proof: Let $I_A$ be the excenter wrt $A$ . Let $|AB|=|AC|=x$ and $\angle BAD=\angle CAD=2\alpha$ . By angle chasing, one can find that $\frac{|AD|}{|AE|}=\frac{1-\sin{(2\alpha)}}{1+\sin{(2\alpha)}}$ and $\frac{|BC|^2}{|DE|^2}=\frac{|BD|^2}{|DI_A|^2}=\frac{\sin^2{(45^\circ-\alpha)}}{\sin^2{(45^\circ+\alpha)}}$ .
You can easily verify that $\frac{1-\sin{(2\alpha)}}{1+\sin{(2\alpha)}}=\frac{\sin^2{(45^\circ-\alpha)}}{\sin^2{(45^\circ+\alpha)}}$ . Hence, $\frac{|AD|}{|AE|}=\frac{|BC|^2}{|DE|^2}$ , as desired. $\;\;\;\;\square$

Now, WLOG $|AC|>|AB|$ .
Let $I_A$ be the excenter wrt $A$ and assume that the tangent at $D$ to the excenter intersects $AB$ and $AC$ at $X$ and $Y$ , respectively.
We know that $|AX|=|AY|$ and excircle of $ABC$ of wrt $A$ is also the excircle of $AXY$ wrt $A$ .
Now, if we apply our claim to $AXY$ , we can find that $\frac{|AD|}{|AE|}=\frac{|XY|^2}{|DE|^2}$ . Hence, it suffices to prove that $\frac{|XY|^2}{|DE|^2}<\frac{|BC|^2}{|DE|^2}\Leftrightarrow |BC|>|XY|$ .
Check that the triangles $ABC$ and $AXY$ have the same perimeter.

Lemma: Given a triangle $ABC$ . We have $\frac{|BC|}{\sin{(\frac A2)}}\ge |AB|+|AC|$ and equality occurs iff $|AB|=|AC|$ .
Proof: Take a point $C'$ on $AB$ such that $|AC'|=|AC|$ and $A\in [BC']$ . We have $\angle AC'C=\frac A2$ . Hence,
$\frac{|BC|}{\sin{(\frac A2)}}=\frac{|BC'|}{\sin{(BCC')}}=\frac{|AB|+|AC|}{\sin{(BCC')}}\ge |AB|+|AC|$ Equality occurs iff $\angle BCC'=90^\circ\Leftrightarrow |AB|=|AC|$ , done.

Thus, we know that $\frac{|BC|}{\sin{(\frac A2)}}>|AB|+|AC|\Leftrightarrow |BC|+\frac{|BC|}{\sin{(\frac A2)}}>|AB|+|AC|+|BC|$ .
By the Law of Sines, one can find that the perimeter of $AXY$ equals to $|XY|+\frac{|XY|}{\sin{(\frac A2)}}$ .
Since the triangles $ABC$ and $AXY$ have the same perimeter, we find that
$|BC|+\frac{|BC|}{\sin{(\frac A2)}}>|XY|+\frac{|XY|}{\sin{(\frac A2)}}\Rightarrow |BC|>|XY|\text{, as desired}.$
Equality occurs iff $|AB|=|AC|$ .

This post has been edited 2 times. Last edited by BarisKoyuncu, Dec 29, 2021, 6:28 PM
Reason: typo

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#5 h Dec 29, 2021, 7:11 PM • 1 Y

Y by GuvercinciHoca

Let $(I_A)$ be tangent to $AC$ at $T$ . Line passing through $D$ cuts $AB,AC$ at $K,L$ . Then $DLI_A \sim DTZ \implies$
$\frac{AD}{AE}=\frac{DT^2}{TZ^2}=\frac{KL^2}{DE^2}\le \frac{BC^2}{DE^2}$

This post has been edited 1 time. Last edited by SerdarBozdag, Dec 29, 2021, 7:11 PM

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#6 h Dec 30, 2021, 9:52 AM • 1 Y

Y by joseph02

Let the tangency point of $A$ excircle and $AB$ be $X$ , then $\frac{AD}{AE}=\frac{AX^2}{AE^2}$ so we need to prove that $\frac{s}{AE} \leq \frac{BC}{2r_a}$ where $s$ is the semiperimeter of $\triangle ABC$ . $s\cdot 2r_a = 4[AXI_A]$ and $BC \cdot AE \geq BC\cdot (AH +2r_a) =2[ABC]+2[I_ABC]=4[AXI_A]$ where $H$ is projecton of $A$ onto $BC$ . This finishes the proof.

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dnztnc

#7 h Jul 24, 2022, 9:49 PM

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Let the center of the excircle be $I$ and the feet of the perpendiculars from $I$ (or tangency points) to $AB$ and $AC$ be $X$ and $Y$ respectively.
And let $\angle BAC = 2a, \angle ABC = 2b, \angle ACB = 2c$ and $|DI|=r$ .
(Note that $a+b+c=90^\circ$ , and $\angle BIX = b$ and $\angle CIY = c$ .)
Finally, we can start the proof.

We need to prove:
$\frac{|BC|^2}{|DE|^2} \ge \frac{|AD|}{|AE|}= \frac{|AD||AE|}{|AE|^2}= \frac{|AX|^2}{|AE|^2}$ which is obviously equivalent to $\frac{|AX|}{|AE|} \le \frac{|BC|}{|DE|}$ .

And:
$\frac{|AX|}{|AE|} \le \frac{|BC|}{|DE|}= \frac{|BX|+|CY|}{|DE|} =\frac{r\tan(b)+r\tan(c)}{2r}=\frac{\tan(b)+\tan(c)}{2}=\frac{\sin(b+c)}{2\cos(b)\cos(c)}$

Also $|AX|=|AI|\cos(a)=|AI|\sin(b+c)$

So we should prove:

$\frac{|AI|\sin(b+c)}{|AE|} \le \frac{\sin(b+c)}{2\cos(b)\cos(c)}$

$\Leftrightarrow 2\cos(b)\cos(c) \le \frac{|AE|}{|AI|}=\frac{|AI|+|IE|}{|AI|}=1+ \frac{r}{|AI|}=1+ \sin(a)= 1+ \cos(b+c)$

$\Leftrightarrow 1 \ge 2\cos(b)\cos(c)-\cos(b+c)=\cos(b-c)$ , which is always true, so we are done.

Equality occurs if and only if we have $b-c=0$ , meaning $|AB|=|AC|$ .

This post has been edited 1 time. Last edited by dnztnc, Jul 24, 2022, 9:56 PM

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#8 h Dec 10, 2022, 6:26 PM

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Let the $A$ -excircle touch $AB$ at $X$ and let $|I_AX| = r_a$ , where $I_A$ is the $A$ -excenter. Note that since the line $AE$ bisects $\angle BAC$ , $I_A$ lies on $AE$ . Also, we have $|AX| = s$ , where $s$ is the semiperimeter of $\triangle ABC$ . Lastly, call $|BC| = a$ .
Now, $|AD| = \sqrt{s^2 + r_a^2} - r_a$ , $|AE| = \sqrt{s^2 + r_a^2} + r_a$ , $|BC| = a$ and $|DE| = 2r_a$ . The requested inequality reduces then to showing that
$\dfrac{\sqrt{s^2 + r_a^2} - r_a}{\sqrt{s^2 + r_a^2} + r_a} \leq \dfrac{a^2}{4r_a^2} \Longleftrightarrow (4r_a^2-a^2)^2(s^2+r_a^2) \leq r_a^2(4r_a^2+a^2)^2 \Longleftrightarrow 4r_a^2(s-a) \leq a^2s$ which since $(s-a)r_a = sr$ ( $r$ is the inradius of $\triangle ABC$ ) simplifies to
$4r_ar \leq a^2$ Finally, as $r_ar = \dfrac{S}{s-a} \cdot \dfrac{S}{s} = (s-b)(s-c)$ ( $S$ is the area of $\triangle ABC$ ),
$4(s-b)(s-c) \leq a^2 \Longleftrightarrow (2s-b-c)^2-4(s-b)(s-c) \geq 0 \Longleftrightarrow (c-b)^2 \geq 0$ This finishes the proof.

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#9 h Dec 10, 2022, 10:17 PM

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This problem is a bit of a ruse: it presents itself as a triangle problem but isn't really about triangles. Sketch:

Fix the angle $A$ and the excircle $\omega_A$ (with center $I_A$ and radius $r_A$ ), and allow $B$ and $C$ to vary such that segment $\overline{BC}$ is tangent to $\omega_A$ . Let $D$ be the tangency point. Then $\angle BI_AD$ and $\angle CI_AD$ are two acute angles which sum to the constant angle $\angle B_IAC = \tfrac{90^\circ + \angle A}2$ , so by Jensen's Inequality $BC = r_A(\tan \angle BI_AD + \angle CI_AD)$ is minimized when both angles are equal to each other. This occurs when triangle $BAC$ is isosceles.

It suffices to show that $BC^2 = DE^2 \cdot\tfrac{AD}{AE}$ in the case where triangle $ABC$ is isosceles, which BarisKoyuncu does in post #4.

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mihaig

#10 h Dec 11, 2022, 10:37 AM

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Great solutions

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