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Source: Iranian third round 2015 geometry problem 5

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824 posts

andria

#1 h Sep 10, 2015, 12:02 PM • 4 Y

Y by buratinogigle, doxuanlong15052000, Adventure10, Mango247

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Let $R$ be the radius of circumcircle of $\triangle ABC$ . Let $A',B',C'$ be the points on $\overrightarrow{AH},\overrightarrow{BH},\overrightarrow{CH}$ respectively such that $AH.AA'=R^2,BH.BB'=R^2,CH.CC'=R^2$ . Prove that $O$ is incenter of $\triangle A'B'C'$ .

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254 posts

A-B-C

#2 h Sep 10, 2015, 1:24 PM • 5 Y

Y by K.N, TheBeatlesVN, Smkh, Adventure10, Mango247

My solution:
Let $D,E,F$ be midpoint of $BC,CA,AB$ .
$A_1,B_1,C_1$ are circumcenters of $\triangle OBC,\triangle OCA,\triangle OAB$ .
Since $\overline{AH}.\overline{AA'}=AO^2$ , $\triangle AHO$ and $\triangle AOA'$ are similar
$\Rightarrow OA'/OH=OA/HA$
Similar ly, $OB'/OH=OB/HB$
$\Rightarrow OA'/OB'=HB/HA=\cos B /\cos C=OE/OD$
$\angle A'OB'=\angle A'OH +\angle B'OH = \angle AHO -\angle AOH +\angle BHO -\angle BOH$
$=360^o -\angle AHB - \angle AOB =180^o +\angle ACB -2\angle ACB = \angle AHB=\angle DOE$
$\Rightarrow \triangle OED \sim \triangle OA'B' \sim \triangle OA_1B_1$
Similarly $\triangle OB_1C_1 \sim \triangle OB'C' ,\triangle OC_1A_1\sim\triangle OC'A'$
$\Longrightarrow A_1B_1C_1O\sim A'B'C'O$
Since $O$ is incenter of $\triangle A_1B_1C_1$ , then $O$ is incenter of $\triangle A'B'C'$
Remark. $\angle A'OB'=\angle AHB=\angle A'HB'$ ...
$\Rightarrow$ $H,O$ are antigonal conjugate WRT $\triangle A'B'C'$
$\Rightarrow$ Nine-point center $N$ of $\triangle ABC$ is Feuerbach point of $\triangle A'B'C'$

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2344 posts

#5 h Sep 10, 2015, 5:47 PM • 4 Y

Y by andria, A-B-C, Adventure10, Mango247

I have an idea

Let $ABC$ be a triangle and $P,Q$ are two isogonal conjugate points. Circle passes through $P,Q$ and is tangent to $QA$ which cuts $PA$ at $D$ . Similarly, we have $E,F$ . Let $XYZ$ be pedal triangle of $P$ .

a) Prove that $\triangle XYZ\cup P\sim\triangle DEF\cup Q$ .

b) Let $R$ be isogonal conjugate of $Q$ with respect to triangle $DEF$ . Prove that $R$ and $P$ are anitgonal conjugate with respect to triangle $DEF$ .

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2312 posts

#8 h Sep 10, 2015, 6:30 PM • 4 Y

Y by buratinogigle, enhanced, Adventure10, Mango247

buratinogigle wrote:

I have an idea

Let $ABC$ be a triangle and $P,Q$ are two isogonal conjugate points. Circle passes through $P,Q$ and is tangent to $QA$ which cuts $PA$ at $D$ . Similarly, we have $E,F$ . Let $XYZ$ be pedal triangle of $P$ .

a) Prove that $\triangle XYZ\cup P\sim\triangle DEF\cup Q$ .

From symmetry, it suffices to prove $\triangle PYZ$ $\stackrel{-}{\sim}$ $\triangle QEF$ . From $\triangle BPQ$ $\sim$ $\triangle BQE$ and $\triangle CPQ$ $\sim$ $\triangle CQF$ we get $\text{[math]}$ $=$ $PQ$ $\cdot$ $(BQ/BP)$ and $FQ$ $=$ $PQ$ $\cdot$ $(CQ/CP)$ , so $EQ/FQ$ $=$ $(CP/BP)$ $\cdot$ $(BQ/CQ)$ . On the other hand, let $\triangle Q_aQ_bQ_c$ be the pedal triangle of $Q$ WRT $\triangle ABC$ , then notice $\triangle BQQ_c$ $\sim$ $\triangle BPX$ and $\triangle CQQ_b$ $\sim$ $\triangle CPX$ we get $(YP/ZP)$ $=$ $(QQ_c/QQ_b)$ $=$ $(CP/BP)$ $\cdot$ $(BQ/CQ)$ $=$ $EQ/FQ$ , so combine $\measuredangle FQE$ $=$ $\measuredangle FQC$ $+$ $\measuredangle CQB$ $+$ $\measuredangle BQE$ $=$ $\measuredangle CPQ$ $+$ $\measuredangle CQB$ $+$ $\measuredangle QPB$ $=$ $\measuredangle CPB$ $+$ $\measuredangle CQB$ $=$ $\measuredangle CAB$ $=$ $\measuredangle YPZ$ we conclude that $\triangle PYZ$ $\stackrel{-}{\sim}$ $\triangle QEF$ .

buratinogigle wrote:

b) Let $R$ be isogonal conjugate of $Q$ with respect to triangle $DEF$ . Prove that $R$ and $P$ are anitgonal conjugate with respect to triangle $DEF$ .

From symmetry, it suffices to prove the reflection $R_d$ of $R$ in $EF$ lie on $\odot (EPF)$ . Since $\measuredangle ER_dF$ $=$ $\measuredangle FRE$ $=$ $\measuredangle DFQ$ $+$ $\measuredangle QED$ $=$ $\measuredangle PZX$ $+$ $\measuredangle XYP$ $=$ $\measuredangle PBC$ $+$ $\measuredangle BCP$ $=$ $\measuredangle BPC$ $=$ $\measuredangle EPF$ , so we get $R_d$ $\in$ $\odot (EPF)$ .

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2312 posts

#10 h Sep 11, 2015, 11:34 AM • 4 Y

Y by mineiraojose, buratinogigle, enhanced, Adventure10

Remark : Here is a nice equivalent property of the generalization mentioned by buratinogigle at post #5 :

Given $\triangle ABC$ and a point $P$ . Let $Q$ be the image of $P$ under the Inversion $\mathbf{I}(\odot (ABC))$ . Let $\triangle P_aP_bP_c,$ $\triangle Q_aQ_bQ_c$ be the pedal triangle of $P,$ $Q$ WRT $\triangle ABC$ , respectively. Let $M$ be the midpoint of $PQ$ and $R$ be the isogonal conjugate of $P$ WRT $\triangle P_aP_bP_c$ . Then $\triangle P_aP_bP_c$ $\cup$ $R$ $\stackrel{-}{\sim}$ $\triangle Q_aQ_bQ_c$ $\cup$ $M$ .

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290 posts

jred

#11 h Oct 21, 2017, 11:06 AM • 2 Y

Y by Adventure10, Mango247

Complex number also works well for this problem，which perfectly handles various configurations.

PS. $\triangle ABC$ should be acute, otherwise $O$ must be one of excenters of $\triangle A'B'C'$ .

This post has been edited 1 time. Last edited by jred, Oct 22, 2017, 4:23 AM

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328 posts

#13 h Aug 6, 2018, 12:48 PM • 2 Y

Y by Adventure10, Mango247

jred wrote:

Complex number also works well for this problem，which perfectly handles various configurations.

PS. $\triangle ABC$ should be acute, otherwise $O$ must be one of excenters of $\triangle A'B'C'$ .

we solve the problem using complex number and we take $\odot ABC$ the unit circle! because $AO^2=AH.AA' \implies \triangle AHO \stackrel{-}{\sim} AA'O \implies \frac{a-0}{a-h}=\frac{\overline{a}-\overline{a'}}{\overline{a}-0} \implies a'=\frac{\sum ab}{b+c}$ and similarly $b'=\frac{\sum ab}{a+c}, c'=\frac{\sum ab}{b+a}$ and now we prove $O$ has the same distance from sides of $\triangle A'B'C'$ in order to prove this we prove the $d(O,AC)$ is symmetric with respect to $a,b,c$ we know that:

$d(O,A'C').A'C'=S(OA'C')$ and with the areal formula it implies:

$\begin{vmatrix} 0 & 0 &1 \\ a' & \overline{a'} &1 \\ c'& \overline{c'} & 1 \\ \end{vmatrix}=|a'-c'|.d(O,A'C')$ and the rest is just 5 minutes of computing!

This post has been edited 2 times. Last edited by H.HAFEZI2000, Aug 6, 2018, 12:50 PM

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