(APF) intersect (AQE)

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(APF) intersect (AQE)

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#1 h Mar 13, 2022, 2:58 AM

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Let $ABC$ be a triangle, and let $BE, CF$ be the altitudes. Let $\ell$ be a line passing through $A$ . Suppose $\ell$ intersect $BE$ at $P$ , and $\ell$ intersect $CF$ at $Q$ . Prove that:

i) If $\ell$ is the $A$ -median, then circles $(APF)$ and $(AQE)$ are tangent.

ii) If $\ell$ is the inner $A$ -angle bisector, suppose $(APF)$ intersect $(AQE)$ again at $R$ , then $AR$ is perpendicular to $\ell$ .

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#2 h Mar 13, 2022, 1:29 PM

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Hint

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#3 h Aug 11, 2023, 8:55 PM

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A solution for the first problem:
Let $S= AB \cap (FPQ)$ , $T= AC \cap (EPQ)$ , $M$ the midpoint of $BC$ and $N = AM \cap ST$ . Then we have that: $AF \cdot AS = AP \cdot AQ = AE \cdot AT$ and hence $SFET$ is cyclic.
So: $\angle AST = \angle AEF = \angle ABC => ST \parallel BC => N$ is the midpoint of $ST => NS=NT (1)$ .
Also from the cyclic quadrilaterals we get that: $\angle SPQ = \angle SFQ = \angle BFC =90 = \angle BEC = \angle PET = \angle MQT$ , hence $SP \parallel QT (2)$ , as they are both perpendicular to $AM$ .
From $(1)$ and $(2)$ we get that $SPTQ$ is a parallelogram. Using the parallelogram and the cyclic quadrilaterals: $\angle AFP = \angle PQS = \angle QPT = \angle QET$ .
Now let $X$ be a point such that $XC<XB$ and $AX$ is tangent to $(AQE)$ .
Then we have that: $\angle AFP = \angle QET = \angle QAE + \angle AQE = \angle QAE + \angle XAE = \angle XAQ = \angle XAP$ , hence $AX$ is also tangent to $(APF)$ , which proves the desired tangency.

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#4 h Aug 12, 2023, 9:54 AM

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For the second problem:
Let $O_1$ be the center of $(AFP)$ , $O_2$ be the center of $(AQE)$ and $H= BE \cap CF$ . By radical axis we have that $O_1O_2 \perp AR$ so it suffices to prove that $O_1O_2 \parallel \ell$ . Since $O_1$ is the circumcenter of $AFP$ we conclude that: $\angle FO_1P = 2 \angle FAP = 2\angle BAP = 2 \angle A/2 = \angle A = 180 - \angle BHC = 180 - \angle FHP$ , hence $FHPO_1$ is cyclic.
Hence, using that $FO_1=PO_1$ , we have that $O_1$ is the midpoint of the arc $FP$ (that does not contain $H$ ) in $(FHP)$ . So $HO_1$ is angle bisector of $\angle FHP$ , and hence angle bisector of $\angle BHC$ , which is obviously parallel to $\ell$ . Hence $HO_1 \parallel \ell$ and analogously we have that $HO_2 \parallel \ell$ . So we have that $O_1O_2 \parallel \ell$ , as needed.

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#5 h Aug 17, 2023, 6:27 AM

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For $\mathrm{i)}$
Let $M$ be the mid-point of $BC$ ,suppose $X,Y\in AM$ satisfy $\angle BXM=90^{\circ}-\angle ACB,\angle CYM=90^{\circ}-\angle ABC$ respectively,
$T$ be the $H$ -Humpty point of $\triangle ABC$
Consider an inversion $\mathcal{I}^{M}_{MB^{2}}$
So $P\mapsto X,Q\mapsto Y,A\mapsto T,F\mapsto F,E\mapsto E$
Hence $\odot(APF)\mapsto \odot(XFT),\odot(AQE)\mapsto \odot(YET)$
We need to prove that $\odot(XFT)$ is tangent to $\odot(YET)\Rightarrow \angle YET+\angle XFT=180^{\circ}$
Note that $H,B,C,T$ are cyclic
Hence $\angle HTC=180^{\circ}-\angle HBC=180^{\circ}-\angle BXM=\angle AXB$
Note that $\angle TBM=\angle BAM$
Hence $\angle THC=\angle BAX$
Which means that $\triangle BAX\sim \triangle CTH$
Similarly $\triangle CYA\sim \triangle BTH$
Consider that $MP\cdot MX=MB^{2}=ME^{2}$
Hence $\angle EXM=\angle BEM=\angle EBM$
Which means that $\angle AXE=180^{\circ}-\angle TXE=180^{\circ}-\angle EBC=180^{\circ}-\angle EAH=\angle HTE$
So $\triangle EAX\sim \triangle EHT$
Similarly $\triangle FYA\sim \triangle FTH$
Since $\dfrac{XB}{YC}=\dfrac{\dfrac{AX\cdot TC}{HT}}{\dfrac{AY\cdot BT}{HT}}=\dfrac{AX}{AY}\cdot \dfrac{TC}{TB}=\dfrac{\dfrac{AX}{HT}}{\dfrac{AY}{HT}}\cdot \dfrac{AC}{AB}=\dfrac{AE}{AF}\cdot \dfrac{HF}{HE}\cdot \dfrac{AC}{AB}=\dfrac{HF}{HE}$
Which means that $\triangle XBF\sim \triangle YCE$
So $\angle YET+\angle XFT=\angle AET+180^{\circ}-\angle YEC+\angle AFT-180^{\circ}+\angle TFB=\angle AET+\angle AFT$
Note that $A,F,H,T,E$ are cyclic
So $\angle YET+\angle XFT=\angle AET+\angle AFT=180^{\circ}$
Which means that $\odot(APF)$ and $\odot(AQE)$ are tangent

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#6 h Aug 17, 2023, 6:33 AM

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For $\mathrm{ii)}$
Let $O_{1},O_{2}$ be the circumcenter of $\triangle AFP$ and $\triangle AEQ$ respectively
Note that $AR$ is the root axis of $\odot(AFP)$ and $\odot(AEQ)$
so we need to prove that $O_{1}O_{2}//AQ$
Consider that $\angle FO_{1}P=\angle BAC=180^{\circ}-\angle FHP$
Which means that $F,O_{1},P,H$ are cyclic
Hence $\angle O_{1}HF=\angle O_{1}PF=90^{\circ}-\dfrac{1}{2}\angle BAC$
Similarly $\angle O_{2}HQ=180^{\circ}-\angle O_{2}EQ=90^{\circ}+\dfrac{1}{2}\angle BAC$
So $\angle O_{1}HF+\angle O_{2}HF=180^{\circ}$
Which means that $O_{1},O_{2},H$ are collinear
Since $\angle O_{1}HF=\angle O_{1}PF=90^{\circ}-\angle FAQ=\angle AQF$
Hence $O_{1}O_{2}//AQ$
Which means that $AR\bot l$

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