PAMO Problem 4: Perpendicular lines

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PAMO Problem 4: Perpendicular lines

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Source: 2019 Pan-African Mathematics Olympiad, Problem 4

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DylanN

#1 h Apr 9, 2019, 6:54 AM • 2 Y

Y by Adventure10, Mango247

The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$ . The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$ . Show that $AO$ is perpendicular to $EF$ .

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pad

#3 h Apr 9, 2019, 7:06 AM • 2 Y

Y by Adventure10, Mango247

We know $AD$ is the $A$ -symmedian of $\triangle ABC$ , so it is well known that $B,C,D,O$ are concyclic. So $(BCD)=(BCO)$ . Then $\angle EAO = \angle CAO=90-B$ , and $\angle AEF=180-\angle CEF=180-\angle CBF=B$ since $E,C,F,B$ are concyclic. Therefore, $AO\perp EF$ .

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DylanN

#4 h Apr 9, 2019, 7:29 AM • 2 Y

Y by Adventure10, Mango247

It's not important that $D$ is defined as the intersection of the tangents. We can replace the circumcircle of $\triangle BCD$ with any circle that passes through $B$ and $C$ . Indeed, we know that $\angle CAO = 90^\circ - \angle B$ , and since $BCEF$ is cyclic, we have that $\angle B = \angle CEF$ , from which the result follows.

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#5 h Apr 9, 2019, 9:35 AM • 1 Y

Y by Adventure10

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#6 h Mar 5, 2024, 12:29 PM

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Observe that we can obtain $\Delta AEF$ by reflecting $\Delta ABC$ about its $A-$ angle bisector and taking a homothety at $A$ . Now since the circumcentre and orthocentre are isogonal conjugates, $O$ has to lie on the $A-$ altitude of $\Delta AEF$ , and therefore we are done. $\square$

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#7 h Mar 6, 2024, 2:06 AM

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Let $H$ be the orthocenter of $ABC$ .
$\measuredangle BEC = \measuredangle BDC = \measuredangle BFC \implies BCEF$ is concyclic $\implies \Delta ABC \sim \Delta AEF$ .
Consider taking a suitable homothety about A then reflecting across the angle bisector of $\angle BAC$ , such that $BC$ goes to $EF$ . Under this transformation, line $AH$ goes to line $AO$ because they are isogonal in $\angle BAC$ . Since $AH \perp BC$ , we have $AO \perp EF$ .

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#8 h Apr 23, 2024, 4:01 PM

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We know that $\measuredangle OAC=90^{\circ}-\measuredangle CBA=\measuredangle KAE$
Also, $\measuredangle CBA=\measuredangle CEF=\measuredangle AEK$
Now, $\measuredangle KAE+\measuredangle AEK+\measuredangle EKA=0$
$\implies 90^{\circ}-\measuredangle CBA+\measuredangle CBA +\measuredangle EKA=0$
$\implies \measuredangle EKA=90^{\circ}$
And we are done!

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fasttrust_12-mn

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fasttrust_12-mn

#9 h Jul 19, 2024, 1:26 AM

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Claim: $O$ lies on the circumcircle of $\triangle BDC$
proof:
$\measuredangle BAC =\measuredangle BCD \implies \measuredangle BDC= 180^{\circ}- 2\times \measuredangle BAC$ , we know that
$\measuredangle BOC = 2\times \measuredangle BAC$ so
$OBDC \text{ is cyclic quad }$
$\measuredangle FBC=\measuredangle AEF$ $\measuredangle OAC=\measuredangle OCA$ $\measuredangle OCB=90^{\circ}$ we know that
$\measuredangle AEG=90^{\circ}-\measuredangle BAO-\measuredangle OAE + \measuredangle BAO\implies 90^{\circ}-\measuredangle OAE$ $\measuredangle AGE =180^{\circ}-(90^{\circ}-\measuredangle OAE+\measuredangle OAE)=90^{\circ}$ hence we are done $\blacksquare$

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#10 h Apr 11, 2025, 12:48 PM

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$\angle EBF=\angle BAF+\angle AFB=\angle A + \angle BDC=\angle A + 180 - 2\angle A = 180 - \angle A$ $\implies FA=FB$ $by\ symmetry\ we \ can\ get \ that\ EA=EC$ hence $F$ lies on perpendicular bisector of $AB$ and $E$ lies on perpendicular bisector of $AC$ moreover $G$ also lies on these lines
$\implies G$ is the orthocentre of $\Delta AEF$ hence $AG$ is perpendicular to $EF$

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#11 h Apr 23, 2025, 11:04 PM • 1 Y

Y by ATM_

To prove that $AO$ perpendicular to $EF$ , it suffies to prove that: $OAC=\frac{\pi}{2}-\measuredangle CFE$
We know that : $\measuredangle OAC=\frac{\pi}{2}-\measuredangle CBA$
And : $\measuredangle CBA=\pi-\measuredangle CBE=\measuredangle CFE$ (because CFEB is cyclic)
$\implies \measuredangle OAC=\frac{\pi}{2}-\measuredangle CFE$
Q.E.D

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ATM_

#12 h Apr 23, 2025, 11:07 PM

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nice solution*

This post has been edited 1 time. Last edited by ATM_, Apr 23, 2025, 11:08 PM
Reason: .

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ATM_

#13 h Apr 23, 2025, 11:10 PM

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here is a cute sketch

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