A line tangent to a circumcircle

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A line tangent to a circumcircle

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Source: Lusophon Mathematical Olympiad 2021 Problem 3

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#1 h Dec 19, 2021, 11:19 PM • 1 Y

Y by jhu08

Let triangle $ABC$ be an acute triangle with $AB\neq AC$ . The bisector of $BC$ intersects the lines $AB$ and $AC$ at points $F$ and $E$ , respectively. The circumcircle of triangle $AEF$ has center $P$ and intersects the circumcircle of triangle $ABC$ at point $D$ with $D$ different to $A$ .

Prove that the line $PD$ is tangent to the circumcircle of triangle $ABC$ .

This post has been edited 1 time. Last edited by ericxyzhu, Jan 1, 2022, 1:54 PM
Reason: Typo

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#2 h Dec 20, 2021, 2:20 AM • 5 Y

Y by jhu08, ericxyzhu, I-love-K2I, Vladimir_Djurica, Kingsbane2139

WLOG $AB>AC$
Since $\angle AEF=90-C$ $\rightarrow$ $\angle PAF = C = \angle ACB$
So $PA$ is tangent to circumcircle of $ABC$
So $PD$ is also tangent to circumcircle of $ABC$ .

This post has been edited 2 times. Last edited by teddy8732, Dec 23, 2021, 2:23 AM
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#3 h Dec 20, 2021, 2:41 AM • 2 Y

Y by jhu08, ericxyzhu

https://cdn.discordapp.com/attachments/919301712685191198/922314477460926485/unknown.png

Consider an inversion around the circumcircle of $\triangle ABC$ .

Let $F '$ be the inverse of $E$ , note that, with $O$ being the center of $(ABC)$ , we have that, as $E$ is in the perpendicular bisector of $BC$ , then $\angle OCE=\angle OAE=\angle OBE \Rightarrow ABOE$ is cyclic, therefore $A,B,F'$ are collinear $\Rightarrow F'=F$ . So we have that $(ABC)$ and $(AEF)$ are orthogonal and the result follows.

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#4 h Dec 20, 2021, 7:57 AM • 2 Y

Y by jhu08, ericxyzhu

Let $O$ be center of $(ABC) \implies \angle APE=2\angle AFE=2(90-\angle ABC) \implies \angle PAC=\angle ABC \implies PA$ tangents to $(O)$ . Since $AD$ is polar of $P$ $\text{wrt}$ $(O)$ , we get $PD$ tangents to $(O)$ ,too.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Dec 20, 2021, 8:19 AM • 1 Y

Y by ericxyzhu

∠AFD = ∠CBD + 90 - ∠B ---> ∠PDA = ∠B - ∠CBD = ∠DBA ---> PD is tangent to circumcircle of triangle ABC.

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#6 h Feb 3, 2025, 7:11 PM

Y by

cute but trivial

Assume all angles are in degrees.

Claim: $PA$ is tangent to $(ABC).$
Proof:
Let $M$ be the midpoint of $BC.$ Extend $PA$ to meet $(AEF)$ at $X\neq A.$ We have that $XA$ is a diameter, so $\angle XEA=90^\circ.$ Then, we have that:
$\angle PAC=\angle PAE=\angle XAE=90-\angle AXE=90-\angle AFE=90^\circ-\angle BFM=\angle FBM=\angle ABC.$ Thus, $\angle PAC=\angle ABC,$ so $PA$ is tangent $(ABC).$ $\blacksquare$

Now, because $PA$ is tangent to $(ABC),$ the other tangent from $P$ to $(ABC)$ must have the same length as $PA.$ The only point $X$ on $(ABC)$ such that $PA=PX$ and $X\neq A$ is $D,$ so we must have that $PD$ is tangent to $(ABC),$ as desired. $\blacksquare$

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#7 h Yesterday at 8:09 PM

Y by

Sneaky but once that is found, trivial angle chase.

Somewhat overly detailed solution (also lacking notation):

LEMMA 1: Proving $PA$ is tangent to the circumcircle of $ABC$ is enough.

PROOF 1: Instead of dealing with the annoying $D$ , we notice that since $A$ and $D$ both lie on the circumcircles of $ABC$ and $AEF$ . Furthermore, $P$ is the center of the circumcircle of $AEF$ . If we can prove that $PA$ is tangent to the circumcircle of $ABC$ , then by symmetry, $PD$ must also be tangent with the circumcircle of $ABC$ . Remove point $D$ for now.

Let $M$ be the midpoint of $BC$ . We start angle chasing. First off, note that $A, B, E$ are collinear so we aim to find $\angle OAB$ first. Let $\angle BAC = a,\angle ABC=b,$ and $\angle ACB = c$ . Therefore $\angle AOB = 2c$ so $\angle OAB = 90-c$ since $OA=OB$ . Now, notice that $\angle AFE = \angle MFC = 90-c$ . Now by the inscribed angle theorem, $\angle APE = 2 \times \angle AFE = 180-2c$ . So $\angle EAP = c$ . Now:
$\angle OAP = 180-\angle OAB -\angle PAE = 180-90+c-c=90$ Therefore $PA$ is tangent to the circumcircle of $ABC$ . Thus line $PD$ is tangent to the circumcircle of triangle $ABC$ .

$\mathbb{Q.E.D.}$

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