IMO 1965 Problem 5

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 7m42s

Problem

Consider $\triangle OAB$ with acute angle $AOB$. Through a point $M \neq O$ perpendiculars are drawn to $OA$ and $OB$, the feet of which are $P$ and $Q$ respectively. The point of intersection of the altitudes of $\triangle OPQ$ is $H$. What is the locus of $H$ if $M$ is permitted to range over (a) the side $AB$, (b) the interior of $\triangle OAB$?

Exploration

The previous attempt failed because it incorrectly assumed a simple affine relation between $M$ and the orthocenter $H$ of $\triangle OPQ$. Testing small cases shows why that cannot hold.

Take $t=1$ (so $OA \perp OB$). Let $M=(1,0)$, so $M$ lies on $OA$. Then $P=M$ and $Q=(\tfrac12,\tfrac12)$. The triangle $OPQ$ has orthocenter $H=(\tfrac12,\tfrac12)$, not $(1,1)$ or any linear scaling of $M$. Thus any formula claiming $H$ is obtained from $M$ by a fixed affine map preserving the first coordinate must fail.

A second test with $M$ varying along $AB$ shows asymmetry: when $M$ approaches $A$, the point $Q$ approaches a fixed point on $OB$, while $P$ moves along $OA$, so $OPQ$ degenerates in a direction that depends on the ratio of distances to the axes. This suggests that $H$ is governed by a symmetric transformation exchanging the roles of $OA$ and $OB$, not a one-sided affine shear.

The correct structure is therefore expected to come from a geometric duality in the right triangle $OPQ$: the orthocenter of a triangle with one vertex at the origin can be expressed using a linear transformation involving the matrix of projection directions. This motivates recomputing $H$ cleanly from perpendicularity conditions rather than trying to reuse any affine guess.

The failure of the previous solution is therefore critical, and the correct approach must restart from the orthogonality conditions defining the orthocenter of $\triangle OPQ$.

Problem Understanding

A point $M$ in the plane determines two projections: $P$ on $OA$ and $Q$ on $OB$. From $P$ and $Q$ we form triangle $OPQ$, and we seek its orthocenter $H$. The goal is to describe the locus of $H$ when $M$ moves along $AB$, and when $M$ moves inside $\triangle OAB$.

The structure is two-stage: first a nonlinear projection map $M \mapsto (P,Q)$, then a second nonlinear construction $ (P,Q) \mapsto H$. The composition is not affine in general, so any solution relying on affine invariance is invalid.

The correct strategy is to express $H$ directly using orthogonality relations in triangle $OPQ$, then eliminate $P,Q$ in terms of $M$.

Key Observations

Since $OPQ$ has vertex $O$ at the origin, the orthocenter can be computed using vector orthogonality relations. If $P=(p,0)$ and $Q=(u,tu)$, then triangle $OPQ$ has sides along coordinate axes only after an oblique change of coordinates aligned with $OB$.

A key simplification is that the orthocenter of a triangle with vertices $O$, $P$, $Q$ satisfies

$$H = P + Q - \pi_{PQ}(O),$$

only in the special case of right triangles, which does not apply here. Thus a direct orthocenter formula must be used.

Instead, we use the defining conditions:

the altitude from $O$ is perpendicular to $PQ$, the altitude from $P$ is perpendicular to $OQ$.

This leads to a linear system in $H$:

$$(H-P)\perp OQ,\qquad (H-Q)\perp OP.$$

These conditions are stable and avoid incorrect geometric shortcuts.

Solution

Let

$$O=(0,0),\quad OA:\ y=0,\quad OB:\ y=tx,\quad t>0.$$

Let $M=(x,y)$. Then

$$P=(x,0),\qquad Q=\left(u,tu\right),\quad u=\frac{x+ty}{1+t^2}.$$

We compute $H=(X,Y)$ as the orthocenter of $\triangle OPQ$.

The condition $(H-P)\perp OQ$ gives

$$(H-P)\cdot (u,tu)=0.$$

Hence

$$(X-x,Y)\cdot (u,tu)=0,$$

$$u(X-x)+tuY=0.$$

Dividing by $u\neq 0$ (the degenerate case $u=0$ corresponds to $M$ on $OA$ and is consistent by continuity),

$$X-x+tY=0.$$

Thus

$$X=x-tY. \tag{1}$$

The second condition $(H-Q)\perp OP$ uses direction $OP=(1,0)$, so altitude from $Q$ is vertical:

$$X=u. \tag{2}$$

Substituting (2) into (1) gives

$$u=x-tY,$$

$$Y=\frac{x-u}{t}.$$

Now substitute $u=\frac{x+ty}{1+t^2}$:

$$Y=\frac{x-\frac{x+ty}{1+t^2}}{t} =\frac{(1+t^2)x-x-ty}{t(1+t^2)} =\frac{t^2x-ty}{t(1+t^2)} =\frac{tx-y}{1+t^2}.$$

Thus

$$H=\left(\frac{x+ty}{1+t^2},\frac{tx-y}{1+t^2}\right).$$

This is the correct orthocenter formula.

Now rewrite in terms of $M$ only:

$$H = A,M,\quad A=\frac{1}{1+t^2} \begin{pmatrix} 1 & t\ t & -1 \end{pmatrix}.$$

This is a linear map with determinant

$$\det A = \frac{-1-t^2}{(1+t^2)^2}=-\frac{1}{1+t^2}\neq 0,$$

so it is invertible.

Therefore the locus problem reduces to applying an invertible linear transformation to the region traced by $M$.

(a) $M$ on segment $AB$

Let

$$A=(a,0),\quad B=(b,tb).$$

Then

$$H(A)=\left(\frac{a}{1+t^2},\frac{ta}{1+t^2}\right),\qquad H(B)=\left(\frac{b(1+t^2)}{1+t^2},\frac{tb-bt^2}{1+t^2}\right)=(b,0).$$

Thus $H(A)=A_1$, $H(B)=B_1$, where

$$A_1=\left(\frac{a}{1+t^2},\frac{ta}{1+t^2}\right),\quad B_1=(b,0).$$

Since the map is linear and invertible, the image of segment $AB$ is the segment $A_1B_1$.

(b) $M$ inside triangle $OAB$

Since the transformation is invertible linear, it maps the interior of triangle $OAB$ bijectively onto the interior of triangle $O A_1 B_1$.

The vertex images are

$$O\mapsto O,\quad A\mapsto A_1,\quad B\mapsto B_1.$$

Thus the locus is the interior of triangle $OA_1B_1$.

Verification of Key Steps

The orthocenter conditions were checked directly from perpendicularity, avoiding any assumption about affine behavior. Substitution of $P$ and $Q$ into the linear system yields a consistent solution for $H$ that satisfies both altitude conditions.

A numerical check with $t=1$, $M=(1,0)$ gives

$$H=\left(\frac12,\frac12\right),$$

matching the direct geometric computation for the right isosceles case.

Linearity is confirmed by writing $H$ explicitly as a matrix transformation of $M$, and invertibility follows from a nonzero determinant.

Both image descriptions are valid because invertible linear maps preserve segments and interiors bijectively.

Alternative Approaches

A synthetic approach uses the fact that $P$ and $Q$ are orthogonal projections onto two rays through $O$. The map $M\mapsto (P,Q)$ can be expressed using orthogonal projection operators, and $H$ arises as the solution to a symmetric orthogonality system in $OPQ$. This leads again to a linear transformation after expressing projections in the basis adapted to $OA$ and $OB$.

Another approach is to rotate coordinates so that $OB$ becomes horizontal; then the problem reduces to symmetric projections onto orthogonal axes followed by a linear shear back to the original frame, producing the same matrix form for $H$.

$$\boxed{ \text{(a) } H \text{ traces the segment } H(A)H(B),\quad \text{(b) } H \text{ fills the triangle } OH(A)H(B) }$$