It is given an acute triangle $ABC$ , $AB \neq AC$ where the feet of altitude from $A$ its $H$. In the extensions of the sides $AB$ and $AC$ (in the direction of $B$ and $C$) we take the points $P$ and $Q$ respectively such that $HP=HQ$ and the points $B,C,P,Q$ are concyclic. Find the ratio $\tfrac{HP}{HA}$.
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Tags: geometry, ratio, circumcircle, geometry proposed
mathuz
19.05.2013 15:48
answer: 1 because $HP=HQ=HA$, $H$ - circumcenter of the triangle $APQ$.
MMEEvN
19.05.2013 15:58
Let $X$ be the point on $AH$ such that $XP \perp AB.\Rightarrow HBPX$ is cyclic.$\Rightarrow AB.AP=AH.AX$ .Since $AB.AP=AC.AQ ,HXQC$ is cyclic..$\Rightarrow XQ\perp AC.\Rightarrow APXQ$ is cyclic..The center of this circle is on $AX$ ..Since $HP=HQ ,H$ is the center of this circle.$\Rightarrow \frac{AH}{HP}=1$
wowmariahaha
10.09.2015 13:03
MMEEvN wrote: Let $X$ be the point on $AH$ such that $XP \perp AB.\Rightarrow HBPX$ is cyclic.$\Rightarrow AB.AP=AH.AX$ .Since $AB.AP=AC.AQ ,HXQC$ is cyclic..$\Rightarrow XQ\perp AC.\Rightarrow APXQ$ is cyclic..The center of this circle is on $AX$ ..Since $HP=HQ ,H$ is the center of this circle.$\Rightarrow \frac{AH}{HP}=1$ How do you know that the center of APXQ is on AX? just curious
parallelines
12.09.2016 19:39
wowmariahaha wrote: MMEEvN wrote: Let $X$ be the point on $AH$ such that $XP \perp AB.\Rightarrow HBPX$ is cyclic.$\Rightarrow AB.AP=AH.AX$ .Since $AB.AP=AC.AQ ,HXQC$ is cyclic..$\Rightarrow XQ\perp AC.\Rightarrow APXQ$ is cyclic..The center of this circle is on $AX$ ..Since $HP=HQ ,H$ is the center of this circle.$\Rightarrow \frac{AH}{HP}=1$ How do you know that the center of APXQ is on AX? just curious Because AP and PX are perpendicular, so AX is diameter in the circle.
Synthetic_Potato
30.05.2018 12:33
My Solution: Notice that the points $P$ and $Q$ are unique (as $AB\neq AC$). So, consider the circle with center $H$ and radius $HA$. Let this circle meet $AB,AC$ at $P',Q'$. It is easy to see that $\angle P'HQ' = 2\angle A$, and $\angle HP'B = 90^\circ - \angle B$. So, $\angle BP'Q' = \angle BCA$. So, $BP'Q'C$ is cyclic and by uniqueness, we are done! $\frac{AH}{HP} = 1$$\blacksquare$.
TFIRSTMGMEDALIST
21.03.2022 01:24
MMEEvN wrote: Let $X$ be the point on $AH$ such that $XP \perp AB.\Rightarrow HBPX$ is cyclic.$\Rightarrow AB.AP=AH.AX$ .Since $AB.AP=AC.AQ ,HXQC$ is cyclic..$\Rightarrow XQ\perp AC.\Rightarrow APXQ$ is cyclic..The center of this circle is on $AX$ ..Since $HP=HQ ,H$ is the center of this circle.$\Rightarrow \frac{AH}{HP}=1$ HP=HQ is not enough to say that H is the circumcenter... is it because AB is diff from AC?
john0512
03.02.2023 19:59
Let $O$ be the circumcenter of $\triangle ABC$. Note that due to BCQP cyclic, reflecting ABC across the A-bisector and then dilating with center A creates triangle $\triangle AQP$. Since $AO$ and $AH$ are isogonal, the circumcenter of $\triangle AQP$ lies on line $AH$. However, due to $HP=HQ$, the circumcenter of $\triangle APQ$ also lies on the perpendicular bisector of $PQ$ (which includes $H$ due to $HP=HQ$), so the circumcenter of $\triangle APQ$ must be $H$, so our answer is 1.
hectorleo123
12.06.2023 04:15
TFIRSTMGMEDALIST wrote: MMEEvN wrote: Let $X$ be the point on $AH$ such that $XP \perp AB.\Rightarrow HBPX$ is cyclic.$\Rightarrow AB.AP=AH.AX$ .Since $AB.AP=AC.AQ ,HXQC$ is cyclic..$\Rightarrow XQ\perp AC.\Rightarrow APXQ$ is cyclic..The center of this circle is on $AX$ ..Since $HP=HQ ,H$ is the center of this circle.$\Rightarrow \frac{AH}{HP}=1$ HP=HQ is not enough to say that H is the circumcenter... is it because AB is diff from AC? Yes, because $2$ lines cannot intersect at more than $1$ point.