Question

Let $ABC$ be an acute-angled triangle with altitude $AK.$ Let $H$ be its ortho-centre and $O$ be its circum-centre. Suppose $KOH$ is an acute-angled triangle and $P$ its circum-centre. Let $Q$ be the reflection of $P$ in the line $HO.$ Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC.$

Answer 1

(a) Consider $f\left(j\right)=a_{m+j}+(1{)}^j{a}_{mj},0\le j\le m$ , where m is a natural number
$f\left(0\right)=2a_m$ is divisible by $2a_m$
Similarly, $f\left(1\right)=a_{m+1}{a}_{m1}=2a_m$ is also divisible by $2a_m$
Assume that $2a_m$ divides $f\left(j\right)$ for all $0\le j<l$ , where $l\le m$ . We prove that $2a_m$ divides $f\left(l\right)$
$f(l-1)=a_{m+l-1}+(-1{)}^{l-1}{a}_{m-l+1}$ ,
$f(l-2)=a_{m+l-2}+(-1{)}^{l-2}{a}_{m-l+2}$ .
Thus we have
$a_{m+l}=2a_{m+l-1}+a_{m+l-2}$
$=2f\left(l1\right)2(1{)}^{l1}{a}_{ml+1}+f(l2\left)\right(1{)}^{l2}{a}_{ml+2}$
$=2f\left(l1\right)+f\left(l2\right)+\left(1{)}^{l1}\right(a_{ml+2}2a_{ml+1})$
$=2f\left(l1\right)+f\left(l2\right)+(1{)}^{l1}{a}_{ml}$ .
This gives
$f\left(l\right)=2f\left(l1\right)+f\left(l2\right)$.
By induction hypothesis $2a_m$ divides $f\left(l1\right)$ and $f\left(l2\right)$. Hence $2a_m$ divides $f\left(l\right)$.We conclude that $2a_m$ divides $f\left(j\right)$ for $0\le j\le m$
(b) We see that $f\left(m\right)=a_{2m}$
Hence, $2a_m$ divides $a_{2m}$ for all natural numbers $m$
Let $n=2^{k}l$ for some $l\ge 1$ . Taking $m=2^{k1}l$ , we see that $2a_m$ divides $a_n$
Using an easy induction , we conclude that $2^k{a}_l$ divides $a_n$
$\therefore 2^k$ divides $a_n$ .