Problem 3

Let

ABC

be an acute triangle with

AB>AC

. Let

\Gamma

be its circumcircle,

H

its orthocenter, and

F

the foot of the altitude from

A

. Let

M

be the midpoint of

BC

. Let

Q

be the point on

\Gamma

such that

\angle HQA=90^\circ

, and let

K

be the point on

\Gamma

such that

\angle HKQ=90^\circ

. Assume that the points

A

B

C

K

Q

are all different and lie on

\Gamma

in this order. Prove that the circumcircles of triangles

KQH

and

FKM

are tangent to each other.

Step 1 of 4: Set up the inversion at

H

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In plain words

The negative inversion centered at the orthocenter that swaps the circumcircle and the nine-point circle is the classical tool for orthocenter configurations.

\iota:\ A\mapsto F,\ Q\mapsto M,\ K\mapsto L

Detailed analysis

Let $L$ be the point on the nine-point circle with $\angle HML=90^\circ$ . The negative inversion $\iota$ centered at $H$ that swaps $\Gamma$ with the nine-point circle sends $A\mapsto F$ (both feet-related points on the corresponding circles through $H$ ), $Q\mapsto M$ (using $\angle HQA=90^\circ$ and $\angle HMA$ -type correspondence for the nine-point circle), and $K\mapsto L$ . Because inversion preserves tangency of circles through the center, and $(KQH)$ passes through the center $H$ of inversion, proving that $(KQH)$ is tangent to $(FKM)$ is equivalent to proving that line $ML$ (the inverse of circle $(KQH)$ , since a circle through the center inverts to a line) is tangent to the inverse of $(FKM)$ , namely circle $(AQL)$ .