Problem 5

Let

\triangle OAB

have acute angle

AOB

. Through a point

M\ne O

, draw perpendiculars to

OA

and

OB

, with feet

P

and

Q

. Let

H

be the orthocenter of

\triangle OPQ

. Find the locus of

H

when

M

ranges over (a) the side

AB

; (b) the interior of

\triangle OAB

Step 3 of 4: Verify the second altitude

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\frac{AM}{MB}=\frac{YQ}{BQ}\\Longrightarrow\frac{YQ}{BQ}=\frac{YH}{HX}\\Longrightarrow QH\\parallel BX\\perp OA.

Detailed analysis

The parallelism $MQ\\parallel AY$ similarly gives $AM/MB=YQ/BQ$ . Combining this with the preceding ratio yields $YQ/BQ=YH/HX$ , hence $QH\\parallel BX$ . Since $BX\\perp OA$ , line $QH$ is the altitude from $Q$ in triangle $OPQ$ .