Mexican midpoint

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Mexican midpoint

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Source: Mexican Mathematical Olympiad 2006 P5

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#1 h Jul 31, 2018, 12:21 PM • 2 Y

Y by GeometryWorld, Adventure10

Let $ABC$ be an acute triangle , with altitudes $AD,BE$ and $CF$ . Circle of diameter $AD$ intersects the sides $AB,AC$ in $M,N$ respevtively. Let $P,Q$ be the intersection points of $AD$ with $EF$ and $MN$ respectively. Show that $Q$ is the midpoint of $PD$ .

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#2 h Apr 1, 2019, 1:56 PM • 1 Y

Y by Adventure10

Note: Though this solution doesn't require $BMNC$ cyclic, still I'll prove it

$\angle AMN=\angle ADN=\angle AFE \implies FE||MN \implies BMNC \text{ is cyclic}$ Let $E'$ be the reflection of $E$ over $AB$ $\implies$ $E' \in \odot (ABDE)$ , then,
$\angle E'FE=2\angle AFE=2\angle BFD=2(90^{\circ}-\angle DFC)=180^{\circ}-\angle DFE$ Hence, $E'F$ passes through $D$ and if $D'$ is the reflection of $D$ over $AB$ $\implies$ $EF$ passes through $D'$ , hence homothety at $D$ with magnitude $\frac{1}{2}$ sends $EF$ $\mapsto$ $MN$ $\implies$ $PQ=QD$

This post has been edited 1 time. Last edited by AlastorMoody, Apr 1, 2019, 2:50 PM

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#3 h Apr 2, 2019, 2:14 AM • 2 Y

Y by Adventure10, Mango247

Alternative Solution:-
Note that $\odot(AEF)$ and $\odot(AMN)$ are tangent at $A$ . Hence the homothety at $A$ which sends $(AEF) \longrightarrow (AMN)$ sends $MN \longrightarrow EF \implies MN \parallel EF$ . Now let $DF \cap MN=J$ and $DE \cap MN=K$ . Note that $\angle FMJ=\angle AFP =\angle AHE = \angle BHD =\angle BFD=\angle MHD \implies JF=JM$ . Now since $\triangle MFJ$ is right angle at $M$ , hence J is the midpoint of $DF$ and similarly $K$ is the midpoint of $DE$ . So there is a homothety centred at $D$ sends $JK \longrightarrow MN$ and it sends $Q \longrightarrow P$ and we are done.

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#4 h Apr 2, 2019, 1:05 PM • 3 Y

Y by DashTheSup, Adventure10, Mango247

Let $M'$ , $N'$ be reflections of $D$ through $AB$ , $AC$ , respectively
We have: $\widehat{BFM'}$ = $\widehat{BFD}$ = $\widehat{ACB}$ = $\widehat{AFE}$
So: $M'$ $\in$ $EF$
Similarly: $N'$ $\in$ $EF$
But: $AD^2$ = $\overline{AM}$ . $\overline{AB}$ = $\overline{AN}$ . $\overline{AC}$ then $B$ , $C$ , $N$ , $M$ lie on a circle
Combine with: $B$ , $C$ , $E$ , $F$ lie on a circle, we have: $EF$ $\parallel$ $MN$
Hence: $\dfrac{\overline{DQ}}{\overline{DP}}$ = $\dfrac{\overline{DM}}{\overline{DM'}}$ = $\dfrac{1}{2}$ or $Q$ is midpoint of $DP$

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#5 h Mar 26, 2020, 1:29 PM

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https://i.imgur.com/VSdrP6d.png

So first of all denote by $\alpha = \angle A$ , $\beta = \angle B$ and $\gamma = \angle C$ ,and denote by $H_k(A)$ a homothety centered at A with homothetic coeficient k.
So in our problem it is easy to see that $\triangle AMD \sim \triangle AFH$ and $\triangle AND \sim \triangle AEH$ ,where $H$ is the orthocenter of the triangle $\triangle ABC$ .Form these similarity relations and from the cyclic quad $AMDN$ we have that $FE \parallel MN$ .Thus we have that $\triangle AMN \sim \triangle AFE$ .Thus we have a $H_k(A)$ ,where $k$ is:
$k = \frac{AH}{AD} =\frac{AB.\frac{cos\alpha}{sin\gamma}}{AB.sin\beta} = \frac{cos\alpha}{sin\beta.sin\gamma}$ Thus we have the following:
$AD-AP=PD$ $\frac{1}{k}AH-AP = PQ+QD$ $\frac{1}{k}(AP+PH)-AP=PQ+QD$ $(\frac{1}{k}-1)AP+\frac{1}{k}PH=PQ+QD$ $(\frac{1}{k}-1)AP=PQ$ Since $\frac{1}{k}PH = QD$ (this is easily seen from the similarities of $\triangle MDN \sim \triangle FHE$ and $\triangle AMD \sim \triangle AFH$ ).
So now we ask ourselves what is the value of $\frac{AD}{AP}$ ,this can be easily calculated:
$\frac{AD}{AP} = \frac{\frac{1}{sin \beta} AM}{\frac{sin \gamma}{sin(90+\beta-\gamma}.AF} = \frac{sin(90+\beta-\gamma)}{sin\beta sin\gamma}\frac{1}{k} = \frac{sin(90+\beta-\gamma)}{cos \alpha}$ Thus we have the following:
$AP+(\frac{1}{k}-1)AP+QD=\frac{sin(90+\beta-\gamma)}{cos\alpha}AP$ Solving this for $QD$ we get that $PQ=QD$ .......

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