Let $ABC$ be an acute triangle with $AB<AC$. Let $D, E,$ and $F$ be the feet of the perpendiculars from $A, B,$ and $C$ to the opposite sides, respectively. Let $P$ be the foot of the perpendicular from $F$ to line $DE$. Line $FP$ and the circumcircle of triangle $BDF$ meet again at $Q$. Show that $\angle PBQ = \angle PAD$.
Problem
Source: PAMO 2023 P6
Tags: geometry
Bluestorm
22.05.2023 23:12
Let $AD$ and $FP$ intersect at $G$. Let $DE$ intersect $AB$ at a point $S$. It stands that $\mathcal{H}(A,F,B,S)$, so $AF/BF = AT/BT$. Since $\angle FPT = 90^{\circ}$, we know that $AP/BP = AF/BF$ so by the internal angle bisector theorem we have that $\angle APG = \angle BPQ \quad ( \star )$. We can now do some angle chasing to find: $\angle AGF = 180^{\circ} - (\angle DAB + \angle AFE + \angle EFP) = 180^{\circ} - (90^{\circ} - \angle B + \angle C + 90^{\circ} - (180^{\circ} - 2\angle B)) =$ $= 180^{\circ} - (90^{\circ} - \angle B + \angle C + 90^{\circ} - 180^{\circ} + 2\angle B) = 180^{\circ} (\angle B + \angle C) =$ $= \angle A$ We also know that $\angle BQF = \angle BDF = \angle A = \angle AGF$, therefore: $\angle PBQ = \angle BQF - \angle BPQ \underbrace{=}_{(\star )} \angle BDF - \angle APG = \angle AGF - \angle APG = \angle PAD$ As desired.
PNT
22.05.2023 23:45
I think the P1 was way better than this. Let $X=DE\cap BA$ then $(B,A;F,X)=-1$ and since $\angle FPE=90$ we get $\angle FPA=\angle BPF$. Now $\angle PAD=180-\angle DPA-\angle PDA=\angle A-(90-\angle EPA)=\angle A-\angle QPB=\angle PBQ$ and done.
Attachments:
fasttrust_12-mn
30.07.2024 20:48
repeated sol>>
emmalearnmath
30.07.2024 20:52
Bluestorm wrote: Let $AD$ and $FP$ intersect at $G$. Let $DE$ intersect $AB$ at a point $S$. It stands that $\mathcal{H}(A,F,B,S)$, so $AF/BF = AT/BT$. Since $\angle FPT = 90^{\circ}$, we know that $AP/BP = AF/BF$ so by the internal angle bisector theorem we have that $\angle APG = \angle BPQ \quad ( \star )$. We can now do some angle chasing to find: $\angle AGF = 180^{\circ} - (\angle DAB + \angle AFE + \angle EFP) = 180^{\circ} - (90^{\circ} - \angle B + \angle C + 90^{\circ} - (180^{\circ} - 2\angle B)) =$ $= 180^{\circ} - (90^{\circ} - \angle B + \angle C + 90^{\circ} - 180^{\circ} + 2\angle B) = 180^{\circ} (\angle B + \angle C) =$ $= \angle A$ We also know that $\angle BQF = \angle BDF = \angle A = \angle AGF$, therefore: $\angle PBQ = \angle BQF - \angle BPQ \underbrace{=}_{(\star )} \angle BDF - \angle APG = \angle AGF - \angle APG = \angle PAD$ As desired. huh........