IMO 1965 Problem 5
Choose Cartesian coordinates with
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 14m52s
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
Choose Cartesian coordinates with
$$O=(0,0),$$
and let the rays $OA$ and $OB$ be the positive coordinate axes. Since $\angle AOB$ is acute, this can always be achieved by a rotation and a positive scaling along the two sides.
Let
$$A=(a,0), \qquad B=(0,b),$$
with $a,b>0$. Let
$$M=(x,y).$$
The feet of the perpendiculars from $M$ to $OA$ and $OB$ are then
$$P=(x,0), \qquad Q=(0,y).$$
The triangle $OPQ$ is right-angled at $O$.
The orthocenter of a right triangle is the vertex at the right angle. Since $OP\perp OQ$, the orthocenter of $\triangle OPQ$ is simply $O$. This would make the locus trivial, which contradicts the nature of the problem. Hence a coordinate choice that identifies $OA$ and $OB$ with coordinate axes must be examined more carefully.
The point $H$ is the orthocenter of $\triangle OPQ$, but $P$ and $Q$ are not arbitrary projections onto two perpendicular lines. The sides $OA$ and $OB$ form an acute angle, not necessarily a right angle. Thus the coordinate axes should instead be chosen along $OA$ and $OB$ as oblique axes.
Let vectors along $OA$ and $OB$ be unit vectors $e_1,e_2$ with
$$e_1\cdot e_2=\cos\theta, \qquad 0<\theta<\frac{\pi}{2}.$$
Write
$$M=ue_1+ve_2.$$
Because $P$ is the orthogonal projection of $M$ onto $OA$,
$$P=(u+v\cos\theta)e_1.$$
Similarly,
$$Q=(u\cos\theta+v)e_2.$$
The natural conjecture is that $H$ depends linearly on $M$. Computing the orthocenter by intersecting two altitudes should reveal an affine transformation $M\mapsto H$.
Carrying out the calculation gives
$$H=v\cos\theta, e_1+u\cos\theta, e_2.$$
Thus $H$ is obtained from $M=ue_1+ve_2$ by exchanging the coefficients and multiplying by $\cos\theta$.
The image of a line segment under an affine map is a line segment, and the image of the interior of a triangle under a nonsingular affine map is the interior of the image triangle. Hence the problem reduces to determining the images of the vertices $O,A,B$.
The most delicate step is the computation of the orthocenter. Any sign error in the altitude equations changes the entire locus.
Problem Understanding
The problem asks for the set of all possible positions of the orthocenter $H$ of the triangle formed by $O$ and the two projection points $P,Q$, where $P$ and $Q$ are obtained by dropping perpendiculars from a moving point $M$ to the sides $OA$ and $OB$ of a fixed acute triangle $OAB$.
This is a Type A problem. A locus must be determined completely. Two things are required. First, every point produced by the proposed locus must actually occur as $H$. Second, every possible orthocenter $H$ must belong to that locus.
The geometric objects involved are orthogonal projections, orthocenters, and affine transformations arising from the coordinates of a point relative to the two sides through $O$.
The difficulty is that $P$ and $Q$ depend on $M$ through orthogonal projection onto nonperpendicular lines. A direct geometric description of $H$ is not apparent. The key idea is to express everything in coordinates adapted to the sides $OA$ and $OB$, then compute the orthocenter explicitly. The resulting map from $M$ to $H$ turns out to be linear.
The answer is that the transformation sends
$$O\mapsto O,\qquad A\mapsto A',\qquad B\mapsto B',$$
where $A'$ lies on $OB$ and $B'$ lies on $OA$, determined by
$$OA' = OA\cos\angle AOB, \qquad OB' = OB\cos\angle AOB.$$
Hence:
For part (a), the locus is the segment $A'B'$.
For part (b), the locus is the interior of the triangle $OA'B'$.
The reason this should be correct is that the map $M\mapsto H$ is affine and nonsingular.
Proof Architecture
Lemma 1. If
$$M=ue_1+ve_2,$$
where $e_1,e_2$ are unit vectors along $OA,OB$, then
$$P=(u+v\cos\theta)e_1, \qquad Q=(u\cos\theta+v)e_2.$$
This follows from the formula for orthogonal projection onto a line.
Lemma 2. The orthocenter of $\triangle OPQ$ is
$$H=v\cos\theta, e_1+u\cos\theta, e_2.$$
This is obtained by writing equations of two altitudes and solving them.
Lemma 3. The map
$$T: M=ue_1+ve_2 \longmapsto H=v\cos\theta, e_1+u\cos\theta, e_2$$
is a nonsingular linear transformation.
Its matrix in the basis $(e_1,e_2)$ is
$$\begin{pmatrix} 0 & \cos\theta\ \cos\theta & 0 \end{pmatrix},$$
whose determinant is $-\cos^2\theta\neq0$.
Lemma 4. Under $T$,
$$O\mapsto O, \qquad A\mapsto A', \qquad B\mapsto B',$$
where
$$OA'=OA\cos\theta, \qquad OB'=OB\cos\theta.$$
This follows by evaluating $T$ on the vertices.
The hardest direction is proving that every point of the proposed locus actually occurs. Lemma 3 resolves this because a nonsingular affine map carries a segment onto a segment and the interior of a triangle onto the interior of its image triangle.
Solution
Let
$$\theta=\angle AOB, \qquad 0<\theta<\frac{\pi}{2}.$$
Choose unit vectors $e_1,e_2$ along the rays $OA$ and $OB$. Then
$$e_1\cdot e_2=\cos\theta.$$
Every point in the plane can be written uniquely in the form
$$M=ue_1+ve_2.$$
Lemma 1
The projections of $M$ onto $OA$ and $OB$ are
$$P=(u+v\cos\theta)e_1, \qquad Q=(u\cos\theta+v)e_2.$$
Proof
Since $P$ is the orthogonal projection of $M$ onto the line generated by $e_1$,
$$P=(M\cdot e_1)e_1.$$
Using
$$M=ue_1+ve_2,$$
and
$$e_1\cdot e_1=1, \qquad e_2\cdot e_1=\cos\theta,$$
we obtain
$$M\cdot e_1=u+v\cos\theta.$$
Hence
$$P=(u+v\cos\theta)e_1.$$
The same argument gives
$$Q=(M\cdot e_2)e_2 =(u\cos\theta+v)e_2.$$
∎
This establishes explicit coordinates for $P$ and $Q$; replacing projection by a coordinate guess would not justify the later altitude computation.
Lemma 2
The orthocenter of $\triangle OPQ$ is
$$H=v\cos\theta, e_1+u\cos\theta, e_2.$$
Proof
Write
$$\alpha=u+v\cos\theta, \qquad \beta=u\cos\theta+v.$$
Then
$$P=\alpha e_1, \qquad Q=\beta e_2.$$
Let
$$H=xe_1+ye_2.$$
Since $OH$ is the altitude through $O$, it is perpendicular to $PQ$.
Now
$$PQ=\beta e_2-\alpha e_1.$$
Hence
$$H\cdot PQ=0,$$
which yields
$$(xe_1+ye_2)\cdot(\beta e_2-\alpha e_1)=0.$$
Expanding,
$$\beta x\cos\theta-\alpha x+\beta y-\alpha y\cos\theta=0.$$
Thus
$$(\alpha-\beta\cos\theta)x = (\beta-\alpha\cos\theta)y.$$
Substituting the expressions for $\alpha,\beta$,
$$u\sin^2\theta, x = v\sin^2\theta, y.$$
Since $\sin\theta\neq0$,
$$ux=vy. \tag{1}$$
Next, the altitude through $P$ is perpendicular to $OQ$, hence
$$(H-P)\cdot Q=0.$$
Using
$$Q=\beta e_2,$$
we obtain
$$(x-\alpha)e_1\cdot e_2+y = 0.$$
Therefore
$$(x-\alpha)\cos\theta+y=0. \tag{2}$$
Substituting
$$\alpha=u+v\cos\theta$$
into (2),
$$y=u\cos\theta+v\cos^2\theta-x\cos\theta. \tag{3}$$
Combining (1) and (3), one verifies that
$$x=v\cos\theta, \qquad y=u\cos\theta$$
satisfy both equations. Since the altitude equations determine a unique intersection point, this is the orthocenter.
Hence
$$H=v\cos\theta, e_1+u\cos\theta, e_2.$$
∎
This identifies the orthocenter explicitly; stopping after writing one altitude equation would not determine the locus.
Lemma 3
The correspondence
$$T(M)=H$$
is a nonsingular linear transformation.
Proof
From Lemma 2,
$$T(ue_1+ve_2) = v\cos\theta, e_1 + u\cos\theta, e_2.$$
In the basis $(e_1,e_2)$, the matrix of $T$ is
$$\begin{pmatrix} 0 & \cos\theta\ \cos\theta & 0 \end{pmatrix}.$$
Its determinant equals
$$-\cos^2\theta.$$
Because $0<\theta<\frac{\pi}{2}$,
$$\cos\theta\neq0,$$
so the determinant is nonzero. Hence $T$ is nonsingular.
∎
This proves that the image of a segment or a triangle is exactly the corresponding segment or triangle image, not merely a subset.
Lemma 4
Let $A'$ and $B'$ be the images of $A$ and $B$ under $T$. Then
$$A'\in OB, \qquad B'\in OA,$$
and
$$OA'=OA\cos\theta, \qquad OB'=OB\cos\theta.$$
Proof
Write
$$A=a e_1, \qquad B=b e_2.$$
Applying $T$,
$$T(A)=a\cos\theta, e_2.$$
Thus $A'$ lies on $OB$ and
$$OA'=a\cos\theta=OA\cos\theta.$$
Similarly,
$$T(B)=b\cos\theta, e_1.$$
Hence $B'$ lies on $OA$ and
$$OB'=b\cos\theta=OB\cos\theta.$$
∎
This identifies the image triangle completely; describing the image only qualitatively would not determine the locus.
For part (a), $M$ ranges over the segment $AB$. Since $T$ is a nonsingular affine transformation, the image of the segment $AB$ is the segment joining the images of its endpoints. Therefore the locus of $H$ is the segment $A'B'$.
For part (b), $M$ ranges over the interior of $\triangle OAB$. Since $T$ is nonsingular, it maps the interior of a triangle onto the interior of the image triangle. The images of the vertices are $O,A',B'$. Hence the locus of $H$ is the interior of $\triangle OA'B'$.
To verify completeness, every point of $A'B'$ arises from a unique point of $AB$, and every point of the interior of $\triangle OA'B'$ arises from a unique point of the interior of $\triangle OAB$, because $T$ is invertible.
Thus the required loci are:
For (a), the segment joining the points $A'$ and $B'$ defined by
$$OA'=OA\cos\angle AOB, \qquad OB'=OB\cos\angle AOB.$$
For (b), the interior of the triangle $OA'B'$.
$$\boxed{\text{(a) locus }=A'B',\qquad \text{(b) locus }=\operatorname{int}(\triangle OA'B')}$$
Verification of Key Steps
The first delicate step is the projection formula. Starting from
$$M=ue_1+ve_2,$$
the projection onto $OA$ equals
$$(M\cdot e_1)e_1.$$
Since
$$M\cdot e_1=u+v\cos\theta,$$
the coefficient of $e_1$ is not merely $u$. Treating $(u,v)$ as rectangular coordinates would produce the incorrect projection and invalidate the remainder of the proof.
The second delicate step is deriving the altitude equation through $O$. Writing
$$PQ=\beta e_2-\alpha e_1,$$
and imposing
$$H\cdot PQ=0,$$
gives
$$ux=vy.$$
A sign error in expanding the dot product would lead to
$$ux=-vy,$$
which changes the image transformation completely.
The third delicate step is the passage from the computed formula for $H$ to the locus. The argument requires that the transformation be invertible. The determinant is
$$-\cos^2\theta.$$
Since the angle is acute,
$$\cos\theta>0.$$
If $\theta=\frac{\pi}{2}$, the determinant would vanish and the image of the triangle would collapse. The hypothesis that the angle is acute is therefore essential.
Alternative Approaches
A synthetic approach is possible. After expressing the projections through similar triangles and right angles, one can show that the coordinates of $H$ with respect to the sides $OA$ and $OB$ are obtained by interchanging the corresponding coordinates of $M$ and multiplying by $\cos\angle AOB$. This leads to the same affine correspondence without introducing vector notation explicitly.
The coordinate method is preferable because the orthocenter is obtained by solving two linear altitude equations. Once the formula
$$H=v\cos\theta, e_1+u\cos\theta, e_2$$
is established, the locus follows immediately from elementary properties of invertible affine transformations.