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 2 of 4: The second intersections B' and C' bisect AY and AX

Previous step Next step

AC'\cdot AC=AY\cdot AM=AX\cdot AN=AX\cdot\tfrac12AC\ \Longrightarrow\ AC'=\tfrac12AX,\quad AB'=\tfrac12AY

Detailed analysis

The real lines $XK$ and $XC$ (the latter is line $AC$ ) already contain points of $\omega_C$ , so each has a second real intersection with $\omega_C$ ; call them $K'$ and $C'$ (a tangent is counted with multiplicity). Define $L',B'$ symmetrically on $\omega_B$ . Using directed products, the secant through $A$ on $\omega_C$ has endpoints $C,C'$ on the $AC$ line and $Y,M$ on the $AY$ line. Thus power from $A$ to $\omega_C$ and to the circle $MYXN$ ; the points $Y,M,C',C$ are endpoints in the preceding secant calculation yields $AC'\cdot AC=AY\cdot AM=AX\cdot AN$ . Since $N$ is the midpoint of $AC$ , $AN=\tfrac12AC$ ; canceling the nonzero directed $AC$ gives $AC'=\tfrac12AX$ , so $C'$ is the midpoint of $AX$ . The symmetric argument gives $B'$ as the midpoint of $AY$ .