AP, BC and OH are concurrent iff AH=HN

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AP, BC and OH are concurrent iff AH=HN

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Source: French TST 2012

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WakeUp

#1 h Aug 2, 2012, 1:07 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle with $AB\not= AC$ . Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$ . $M$ is the midpoint of $BC$ . The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$ . Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$ .

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#2 h Aug 2, 2012, 1:54 PM • 2 Y

Y by Adventure10 and 1 other user

My solution is based on long calculation. So I'm giving outline:
Let the intersection point be $D$ . We'll easily get $BD$ in terms of $a,b,c$ .
So we'll also get $AD^2$ using cosine rule.
We've $PD^2 \cdot AD^2=CD^2 \cdot BD^2.$ Hence we'll get $PD^2.$
Also we've $AP^2-PD^2=AM^2-MD^2.$ Hence we've $AP^2$
Using $AD=AP+PD$ we'll get a equation in terms of $a,b,c$ .
From here it's easy to show the iff part.

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#3 h Aug 2, 2012, 2:45 PM • 3 Y

Y by YNJTheGreat, Adventure10, Mango247

Let $K$ be the antipode of $A$ WRT $\Gamma.$ Since $KC,BH$ are both perpendicular to $AC$ and $KB,CH$ are both perpendicular to $AB,$ then $HCKB$ is parallelogram $\Longrightarrow$ $K$ is the reflection of $H$ about $M.$ Thus, $\angle APM=90^{\circ}$ implies that $M,H,P$ are collinear. If $D \equiv AP \cap BC,$ then $H$ is also the orthocenter of $\triangle ADM.$ Consequently, $OH \in D$ $\Longleftrightarrow$ $OH \perp AM$ $\Longleftrightarrow$ $OH$ is perpendicular bisector of $\overline{AN}$ $\Longleftrightarrow$ $AH=HN.$

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r1234

#4 h Aug 3, 2012, 6:10 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

In other way....
Let $(M)$ be the circle with diameter $BC$ .Now let $\Gamma$ be the circle with diameter $AM$ .Then $AP$ is the radical axis of $(O)$ and $\Gamma$ . Again $BC$ is the radicla axis of $(O)$ and $(M)$ . Now note that $H$ has the sam e power w.r.t $\Gamma$ and $(M)$ . So we just need to prove that $O$ lies on the radical axis of $\Gamma$ and $(M)$ .But centre of $\Gamma$ lies on $AM$ .Let $OH\cap BC=K$ . Then $KOH$ is the radical axis of $\Gamma$ and $(M)$ . So $OH\perp AN\Longrightarrow AH=HN$ . The only if part can be proven in the same way.

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#5 h Aug 3, 2012, 1:33 PM • 3 Y

Y by YNJTheGreat, Adventure10, Mango247

The case when $AB<AC$ is similar with $AB>AC$ , then suppose that $AB<AC$ . Let $HM$ intersect $\Gamma$ at $Q$ and $R$ , with $Q$ on the arc $AB$ and $R$ on the arc $BC$ . Then we claim that $\angle AQR=90$ . To prove this, suppose that $HM$ intersect $AO$ at point $R'$ . Note that $AH||OM$ , and $AH=2OM$ . Then we get that $\Delta R'MO$ is similar with $\Delta R'AH$ with the ratio of similitude 2. Then $R'A=2OA$ and $R',O,A$ are collinear $\rightarrow$ $R'A$ is the diameter of $\Gamma$ , which implies that $R'$ lies on $\Gamma$ , and finally $R'=R$ (since $H,M,R'$ are collinear too). Then $RA$ is the diameter of $\Gamma$ , and our claim was proven.

We see that $\angle AQR=90\rightarrow\angle AQM=90$ . Then $Q$ lies on a circle with diameter $AM$ , and finally the circle with diameter $AM$ cuts $\Gamma$ at $A$ and $Q$ , which directly implies that $Q=P$ . Now let $AP$ intersect $BC$ at point $X$ . Consider the triangle $XAM$ . Note that $AH\bot BC\rightarrow AH\bot XM$ , and $MH\bot AP\rightarrow MH\bot AX$ . Then $H$ is the orthocentre of $\Delta XAM$ , and we get that $XH\bot AM$ .

Note that $AH=HN$ is equivalent with $H$ lies on the perpendicular bisector of $AN$ . Since $O$ is obviously lies on the perpendicular bisector of $AN$ , we get that the condition is equivalent with $HO\bot AN\Leftrightarrow HO\bot AM$ . We have known that $XH\bot AM$ . If $HO\bot AM$ , then $X,H,O$ are collinear $\Leftrightarrow$ $AP,BC,OH$ meet at a common point. Then we have that $AP,BC,OH$ are concurrent iff $AH=HN$ (Q.E.D)

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bobthesmartypants

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bobthesmartypants

#6 h Feb 27, 2017, 6:03 PM • 2 Y

Y by Adventure10, Mango247

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