The line passes through the incenter

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Source: Croatia TST 2009

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213 posts

djuro

#1 h Apr 15, 2009, 2:07 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

A triangle $ABC$ is given with $\left|AB\right| > \left|AC\right|$ . Line $l$ tangents in a point $A$ the circumcirle of $ABC$ . A circle centered in $A$ with radius $\left|AC\right|$ cuts $AB$ in the point $D$ and the line $l$ in points $E, F$ (such that $C$ and $E$ are in the same halfplane with respect to $AB$ ). Prove that the line $DE$ passes through the incenter of $ABC$ .

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4148 posts

#2 h Apr 15, 2009, 2:45 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$I$ is the incenter and $DI$ cuts $\ell$ at $E_0.$ Since $\angle BIC=\angle BDC=90^{\circ}+\frac{_1}{^2}\angle BAC,$ it follows that $B,I,C,D$ are concyclic $\Longrightarrow$ $\angle E_0IC=\angle ABC.$ Then $\angle ABC=\angle E_0AC$ $\Longrightarrow$ $A,I,C,E_0$ are concyclic. Since $\angle AID=\angle AIC,$ then chords $AC=AE_0$ are equal $\Longrightarrow$ $E \equiv E_0,$ as desired.

Remark: Similarly, we show that $DF$ passes through the A-excenter of $\triangle ABC.$

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46 posts

#3 h Apr 20, 2009, 3:20 AM • 2 Y

Y by Adventure10 and 1 other user

Very easy. Call $I$ is the incenter of $\Delta ABC$ . We have:
$\angle ADE=\pi-frac{\angle A+ \angle B}{2}=frac{\angle C}{2}=\\angle ACI=angle ADI=$
Hence we get the result.

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49 posts

#4 h Apr 20, 2009, 1:03 PM • 4 Y

Y by Adventure10 and 3 other users

In fact Where I first met this problem was in China's National High-school Math Competition(I don't konw the official English name of this contest, I just translated the words) in year 2005.
It was in the second test.The second test contains 3 olympiad-type problems, each problem worths 50 points.120 minutes are allowed in the second test.
I did this problem easily even though I was still a Junior student at that time.

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7054 posts

#5 h Apr 20, 2009, 3:20 PM • 2 Y

Y by Adventure10 and 1 other user

djuro wrote:

Let $\triangle ABC$ with the incenter $I$ and $c>b$ . Denote $D\in (AB)$ for which $AD=b$ and $E\in AA$ (the tangent in a point $A$

to the circumcircle of $\triangle ABC$ ) so that $AE=b$ and the sideline $AB$ doesn't separate $C$ , $E$ . Prove that the line $I\in DE$ .

Proof. $X\in AI\cap DE$ and $Y\in BC\cap AI\ \implies\ m(\angle BAE)=A+B$ , $YD=YC=\frac {ab}{b+c}$ , $m(\angle ADY)=C$ .

Thus, $DY\parallel AE\ \implies\ \frac {XA}{XY}=\frac {AE}{DY}=$ $\frac {b}{\frac {ab}{b+c}}\ \implies\ \frac {XA}{XY}=\frac {b+c}{a}$ $\stackrel{\mathrm{Van\ Aubel}}{\ \ \implies\ \ }X: =I$ . In conclusion, $I\in DE$ .

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429 posts

#6 h May 6, 2011, 6:32 AM • 1 Y

Y by Adventure10

Denote $\widehat{BAC}=2a,\widehat{ACB}=2c,\widehat{ABC}=2b$ (after angle chasing) we have $\widehat{EAC}=2b$ (using $\triangle{ACE},\triangle{ACD}$ are isosceles we have $\widehat{ADE}=90-a-b$ ,so $\widehat{EDC}=b,\widehat{ACI}=c,\widehat{ICD}=90+a-c$ and all we have to do is to aplt Ceva's reciprocal theorem in tre trigonometric form. The concluision follows only using thing like ${sin a sin b=-\frac{1}{2}(cos(a+b)-cos(a-b)}$

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4 posts

#7 h May 10, 2011, 11:00 PM • 1 Y

Y by Adventure10

Let $\angle CAB=2\alpha$ , $\angle ABC=2\beta$ , $\angle ACB=2\gamma$ . Suppose that $G\in BC$ such that $AG$ is bisector of $\angle A$ . Let $I$ be the intersection point of $DE$ with $AG$ . By hypothesis $\angle EAC=2\beta$ , also the triangles $ADC$ , $ADE$ and $ACE$ are isosceles, therefore making some calculations, $\angle AED=\gamma$ , $\angle EDC=\alpha$ and $\angle IAC=\alpha$ , hence AICE is cycle and $\angle ACI=\angle AEI=\angle AED=\gamma$ , then $CI$ is bisector of $\angle C$ thus $I$ is the incenter.

[geogebra]5e812a8a2647e28f7ba0108b9e11c654a4a76154[/geogebra]

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761 posts

#8 h Nov 13, 2011, 5:09 AM • 2 Y

Y by Adventure10, Mango247

linboll wrote:

In fact Where I first met this problem was in China's National High-school Math Competition(I don't konw the official English name of this contest, I just translated the words) in year 2005.
It was in the second test.The second test contains 3 olympiad-type problems, each problem worths 50 points.120 minutes are allowed in the second test.
I did this problem easily even though I was still a Junior student at that time.

yes,it is.and it can be second-killed by just some angle substitutions.
by the way,the official name is China Second Round.

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118 posts

#9 h Nov 24, 2011, 6:57 PM • 2 Y

Y by Adventure10, Mango247

It can be also easily done with $S_{\triangle ADE}=S_{\triangle AID}+S_{\triangle AIE}$

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Math_Is_Fun_101

159 posts

Math_Is_Fun_101

#10 h May 9, 2022, 12:58 AM

Y by

Let $I$ be the intersection of the interior angle bisector with $\overline{DE}$ . We have $\angle CEI = \angle CED = \frac12\angle CAD = \frac12\angle A = \angle CAI$ , so $AICE$ is cyclic. Then, $\angle ICA = \angle IEA = \angle DEA = \frac12\angle DAF = \frac12\angle C$ , so $\overline{IC}$ bisects $\angle C$ . Hence, $I$ lies in two interior angle bisectors, so it is the incenter. $\blacksquare$

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69 posts

#11 h Apr 28, 2025, 3:54 PM

Y by

Note that $AI$ is the angle bisector of $\angle A$ and $\Delta ADC$ is isosceles $\implies$ $AI$ is perpendicular bisector of $DC$
so $\angle ADI = \angle ACI = \angle ICB$ $\implies$ $ADIC$ is cyclic and hence $\angle FDE = 90+ \angle IBC = \angle AIC$ so $AICE$ is cyclic hence $\angle ADI = \angle ICA = \angle IEA$ $\implies D-I-E$

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