Tangents intersect on AC

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Source: All Russian 2017,grade 10,day 2,P8

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237 posts

#1 h May 1, 2017, 5:47 AM • 4 Y

Y by doxuanlong15052000, Adventure10, Mango247, NO_SQUARES

In a non-isosceles triangle $ABC$ , $O$ and $I$ are circumcenter and incenter,respectively. $B^\prime$ is reflection of $B$ with respect to $OI$ and lies inside the angle $ABI$ .Prove that the tangents to circumcirle of $\triangle BB^\prime I$ at $B^\prime$ , $I$ intersect on $AC$ . (A. Kuznetsov)

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Lsway

#2 h May 1, 2017, 6:12 AM • 5 Y

Y by mhq, doxuanlong15052000, like123, Adventure10, NO_SQUARES

This problem is too simple ,as this structure has appeared before.
solution
Here is Chinese translation of geometry problems in 2017 ARMO:
ARMO几何试题的中文译文

This post has been edited 3 times. Last edited by Lsway, May 11, 2017, 4:39 AM
Reason: thanks jred for pointing out the typo

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#3 h May 1, 2017, 2:21 PM • 2 Y

Y by Adventure10, Mango247

Aha! Nice problem!

Let $L$ be a point on $BC$ such that $\angle OIL=90^{\circ}$ . It's clear that $IL$ is tangent to $(BIB')$ so it suffices to check $LI=LB'$ . To see this actually holds, let $M$ be the point on the external bisector of angle $ABC$ such that $M, I, L$ are collinear.

Claim: $IM=2IL$ .

(Proof) Let $A', C', L'$ be points symmetric to $I$ in $A, C, L$ respectively; $I_A, I_B, I_C$ be the excenters opposite to $A, B, C$ in $\triangle ABC$ . Let $IL$ meet the circumcircle of triangle $I_AI_BI_C$ at $X, Y$ . Note that $I$ is the midpoint of $XY$ and $A', C'$ lie on this circle so by Butterfly's Theorem we get $IM=IL'$ as claimed. $\square$

Evidently, $IM=2IL$ so the midpoint of $IM$ is equidistant from $B$ and $I$ . Reflecting in $OI$ yields the conclusion.

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Gems98

#4 h May 3, 2017, 6:43 AM • 1 Y

Y by Adventure10

How to use butterfly? I don't understand.

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DSD

#5 h May 8, 2017, 6:17 AM • 1 Y

Y by Adventure10

Well... This configuration is so popular that this is a bad problem for olympiad.

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290 posts

jred

#6 h May 11, 2017, 2:57 AM • 2 Y

Y by Adventure10, Mango247

Lsway wrote:

This problem is too simple ,as this structure has appeared before.
Let $\{ D,A\}=AI \cap \odot O,\{ E,C\}=AI \cap \odot O.$ The line through $I$ perpendicular to $OI$ intersects $AC,DE$ at $F,G.$
Obviously $IG=IB$ .From Butterfly Theorem we can know $IG=IF$ .Hence $GBB^\prime F$ is a isosceles trapezoid $\Longrightarrow IF=B^\prime F$
$OI$ passes through the circumcenter of $\triangle BB^\prime I$ $\Longrightarrow$ $IF$ is tangent to $\odot (BB^\prime I).$
$\therefore F$ is the pole of $B^\prime I$ WRT $\odot (BB^\prime I)$

There's a typo in " $IG=IB$ ", which should be $IG=BG$ .

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Pqrq

#9 h Feb 4, 2021, 7:22 PM • 4 Y

Y by guptaamitu1, sabkx, Mango247, Mango247

A synthetic solution:
First, we note that $B'$ lies on $(ABC)$ , and the tangent line from $I$ to $(BIB')$ is perpendicular to $OI$ . Let this line intersect line $AC$ at point $D$ . Let $(B'ID)$ intersect $(ABC)$ again at $E$ and let $BI$ intersect $(ABC)$ at $F$
Claim: $E,D$ and $F$ are collinear.
Proof: From the cyclic quadrilaterals, we have $\angle(B'EF)$ $=$ $\angle(B'BI)$ $=$ $\angle(B'ID)$ $=$ $\angle(B'ED)$ (the angles are directed)
which means that $E,D$ and $F$ are on the same line.
On the other hand, it is well known that $FI^2$ $=$ $FA^2$ $=$ $FD.FE$ (just consider that the triangles $FEI$ and $FID$ are similar from some angle chasing), which gives us that $\angle(DIF)$ $=$ $\angle(DEI)$ $=$ $\angle(DB'I)$ . As the line $IO$ is perpendicular to the line $BB'$ and $ID$ , we get that $BB' \parallel IO$ $\Rightarrow$ $\angle(B'BI)$ $=$ $\angle(DIF)$ $\Rightarrow$ $\angle(DB'I)$ $=$ $\angle(B'BI)$ , which gives us the desired result.

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308 posts

#10 h Oct 7, 2021, 5:00 AM

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Let the tangent of $(BIB')$ at $B$ meet $AC$ at $N$ . Let $I'$ be the $B$ -excentre of $\triangle ABC$

Claim. $I'N$ is parallel to $BB'$
Proof.
Consider the $\sqrt{ac}$ inversion at $B$ together with a reflection w.r.t. the angle bisector of $\angle ABC$ . Then $I$ and $I'$ are interchanged by a well-known fact, and since
$\angle NBI=\angle BB'I=\angle IBB'$ $N,B'$ are interchanged. Therefore, let $B"$ be the reflection of $B'$ w.r.t. $BI'$ , then
$BN\times BB"=BI\times BI'$ so $\angle NI'I=\angle NB"I=\angle BB'I=\angle IBB'$ as desired.

Let $AC$ and $NI$ meet $BB'$ at $Y$ and $Z$ , then notice that
$(Z,\infty_{BB'};B,Y)\overset{N}{=}(I,I';B,X)=-1$ hence $Z$ is the midpoint of $BY$ . Let the tangent at $I$ to $(BB'I)$ meet $AC$ at $K$ . By Menelaus Theorem,
$\frac{XN}{NY}=\frac{XI}{IB}=\frac{XK}{KY}$ Hence $(N,K;X,Y)=-1$ , let $BK$ meet $(BB'I)$ at $M$ , then
$(B,M;I,B')\overset{B}{=}(N,K;X,Y)=-1$ so $BM$ is the symmedian, and hence $B'K$ is also tangent to $(BB'I)$ as desired.

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#11 h Jan 1, 2024, 1:30 PM

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Let $J = (B'ID) \cap (ABC)$ , Let $E$ be the arc midpoint; Let $D$ be on $AC$ such that $\measuredangle OID = 90^\circ$ .
Claim: $E-J-D$
Proof
By Shooting lemma and Incenter-excenter lemma, $MI^2=MA^2=MJ \cdot MD$ ; $\measuredangle DB'I = \measuredangle DJI = \measuredangle EID$ which is equal to $\measuredangle DIB'$ due to reflection, Hence we're done.

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#12 h Mar 24, 2024, 5:49 PM • 1 Y

Y by buratinogigle

Since $OB=OB'$ we have $B' \in (ABC)$ .
Let $K$ be a point of intersection tangents to $(BB'I)$ in points $B'$ and $I$ . We want to prove that $K \in AC$ .
Note that since $BI=B'I$ , we have $\Delta BB'I \sim \Delta IB'K$ .
Now let's use complex coordinates.
Let $(ABC)$ be a unit circle, $A$ has coordinate $a^2$ , $C$ has coordinate $\frac{1}{a^2}$ and $B$ has coordinate $b^2$ so that $I$ has coordinate $-(1+ab+\frac{b}{a})=j$ . Also let $B'$ has coordinate $x \not = b^2$ and $K$ has coordinate $k$ .

We have $(j-b^2)\overline{(j-b^2)}=(j-x)\overline{(j-x)} \Rightarrow \frac{j}{b^2}+b^2\overline{j}=\frac{j}{x}+x\overline{j} \Rightarrow j\frac{x-b^2}{b^2x}=\overline{j}(x-b^2) \Rightarrow j=b^2x\overline{j}$ $\Rightarrow x=\frac{j}{b^2\overline{j}}=\frac{-(1+ab+b/a)}{-b^2(1+\frac{1}{ab}+a/b)}=\frac{a+a^2b+b}{ab^2+a^2b+b}.$ Now since $\Delta BB'I \sim \Delta IB'K$ , we get $\frac{b^2-x}{j-x}=\frac{j-x}{x-k}$ $\Rightarrow k=\frac{j^2-2jx+b^2x}{b^2-x}=\frac{(1+ab+b/a)^2+2(1+ab+b/a)\frac{a+a^2b+b}{ab^2+a^2b+b}+b^2\frac{a+a^2b+b}{ab^2+a^2b+b}}{b^2-\frac{a+a^2b+b}{ab^2+a^2b+b}}=\frac{(a^2b+a+b)(a^2b+2a+b)}{a^2(b-1)(b+1)}.$ And we want to check that $k+\overline{k}=a^2+1/a^2$ .
Note that $\overline{k}=\frac{-(a^2+ab+1)(a^2+2ab+1)}{a^2(b-1)(b+1)}$ so $k+\overline{k}=\frac{(a^2b+a+b)(a^2b+2a+b)-(a^2+ab+1)(a^2+2ab+1)}{a^2(b-1)(b+1)}=\frac{a^4b^2+b^2-a^4-1}{a^2(b^2-1)}=\frac{(b^2-1)(a^4+1)}{a^2(b^2-1)}=\frac{a^4+1}{a^2}=a^2+1/a^2$ and it is end of solution!

This post has been edited 9 times. Last edited by NO_SQUARES, Mar 25, 2024, 8:42 AM

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#13 h Feb 14, 2025, 9:29 AM

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Invert around the incircle.
New Problem Statement: $ABC$ is a triangle with circumcenter, orthocenter $O,H$ respectively. $M$ is the midpoint of $BC$ and $N$ is the reflection of $M$ over $OH$ . $K$ lies on the circle with diameter $AO$ and $KO\perp OH$ . Prove that $(ONK)$ is tangent to $MN$ .
It sufficies to show that $NK=NO$ . Let $V$ be the altitude from $A$ to $OH$ . Let's show that $AVMN$ is a parallelogram. Let $A'$ be the antipode of $A$ and $W$ be the foot of the altitude from $A'$ to $OH$ . Since $OA=OA'$ and $HM=MA'$ , we have $AV=A'W=MN$ as desired. $\blacksquare$

This post has been edited 1 time. Last edited by bin_sherlo, Feb 14, 2025, 9:41 AM

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