$O$ is circumcircle and $AH$ is the altitude of $\triangle ABC$. $P$ is the point on line $OC$ such that $AP \perp OC$. Prove, that midpoint of $AB$ lies on the line $HP$.
Problem
Source: Moscow Olympiad 2018, Grade 10, P3
Tags: geometry, circumcircle
PavelMath
13.07.2018 15:32
RagvaloD wrote: $O$ is circumcircle and $AH$ is the altitude of $\triangle ABC$. $P$ is the point on line $OC$ such that $AP \perp OC$. Prove, that midpoint of $AB$ lies on the line $HP$. This is a well-known problem. Let $M$ be the intersection point of $AB$ and $HP$. We get: $$\angle PHA=\angle PCA=\angle CAO=\angle MAH, $$so $AM=MH$ and $MH=BM$ (because $\angle BHA=90^\circ$). Hence $AM=BM$.
MAKEANALITGREATAGAIN2018
13.07.2018 15:38
Let $A=(0,0) , H=(2,0) , B = (2,2b) , C = (2,2c)$. It is easy to prove that $O = (1-bc,b+c)$ and $M = (1,b)$. Line $OC$ is $y = \frac{c-b}{1+bc} x + \frac{2b(c^2+1)}{1+bc}$ and line $AP$ is $y = \frac{1+bc}{b-c}x$. Intersecting them gives $P = (\frac{2b(b-c)}{b^2+1},\frac{2b(1+bc)}{b^2+1})$. The slope of $PH$ is $\frac{\frac{2b(1+bc)}{b^2+1}-0}{\frac{2b(b-c)}{b^2+1}-2} = -b$, which is also the gradient of $HM$ .
