Let $ABC$ be an acute triangle. Let$H$ and $D$ be points on $[AC]$ and $[BC]$, respectively, such that $BH \perp AC$ and $HD \perp BC$. Let $O_1$ be the circumcenter of $\triangle ABH$, and $O_2$ be the circumcenter of $\triangle BHD$, and $O_3$ be the circumcenter of $\triangle HDC$. Find the ratio of area of $\triangle O_1O_2O_3$ and $\triangle ABH$.
Problem
Source:
Tags: geometry, circumcircle, ratio
tk1
21.01.2013 05:27
Note that $O_1,$ $O_2,$ and $O_3$ are the midpoints of $AB,BH,$ and $HC,$ respectively. Thus, $O_1O_2\perp BH$ and $O_1O_2=AH/2.$ Also, $O_2O_3\parallel BC$ and $O_2O_3=BC/2.\ (1)$ Extend $O_1O_2$ until it intersects $AB$ at $E.$ Then $\triangle EO_1O_2$ and $\triangle BAC$ are similar. If we denote by $h_1,h_A$ the heights from $O_1,$ $A,$ respectively, we get $h_1/h_A=O_1O_2/AC=(1/2)\cdot AH/AC.\ (2)$ In addition, $(O_1O_2O_3)=(1/2)\cdot h_1\cdot O_2O_3,$ and $(ABH)=(1/2)\cdot BH\cdot AH.\ (3)$ Substituting $(1),(2)$ in $(3),$ we get $(O_1O_2O_3)/(ABH)=1/4.$
lckihi
04.10.2018 11:27
Note that $O_1$ and $O_2$ are midpoints of $AB$ and $BH$ respectively. Then $[O_1O_2H]:[ABH]=1:4$. But by midpoint theorem, $O_1O_2//AH$. So $[O_1O_2H]=[O_1O_2O_3]$. So the desired ratio is $1:4$.
