Problem 2

Let

ABC

be a triangle and let points

M

and

N

be the midpoints of sides

AB

and

AC

, respectively. Let points

K

and

L

be chosen strictly inside triangles

BMC

and

BNC

, respectively, such that

K

lies strictly inside triangle

ABL

and

L

lies strictly inside triangle

AKC

. Suppose that

\angle KBA=\angle ACL

\angle LBK=\angle LNC

, and

\angle LCK=\angle BMK

. Let

O

be the circumcentre of triangle

AKL

. Prove that

OM=ON

Step 4 of 4: M and N have equal power to (AEF), forcing OM = ON

Previous step Next step

MA\cdot ME=\tfrac12AM\cdot AY=\tfrac12AN\cdot AX=NA\cdot NF\ \Longrightarrow\ OM^2-R^2=ON^2-R^2\ \Longrightarrow\ OM=ON

Detailed analysis

Since $M$ is the midpoint of $AB$ and $E$ is the midpoint of $BY$ , $ME=\tfrac12AY$ (directed along line $AB$ ); similarly $NF=\tfrac12AX$ along line $AC$ . Using $AM\cdot AY=AN\cdot AX$ from the cyclic quadrilateral $MYXN$ (step 1), the powers of $M$ and $N$ with respect to circle $(AEF)$ satisfy $MA\cdot ME=\tfrac12AM\cdot AY=\tfrac12AN\cdot AX=NA\cdot NF$ . Since $O$ is the centre of $(AKL)=(AEF)$ and $R$ its radius, the power of a point equals $d^2-R^2$ , so $OM^2-R^2=ON^2-R^2$ , giving $OM=ON$ .