Kvant Math Problem 64

Let $A$ and $B$ be the feet of the altitudes from $Q$ and $P$ onto the sides $PM$ and $QM$ respectively.

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Problem

On the plane, a line $l$ and two points $P$ and $Q$ lying on the same side of it are given. Find a point $M$ on the line $l$ such that the distance between the feet of the altitudes of triangle $PQM$ dropped onto the sides $PM$ and $QM$ is minimized.

G. A. Palatnik

Exploration

Let $A$ and $B$ be the feet of the altitudes from $Q$ and $P$ onto the sides $PM$ and $QM$ respectively. The quantity to be minimized is $AB$.

A direct coordinate computation seems natural. Put $M$ at the origin and let

$$P=\mathbf p,\qquad Q=\mathbf q.$$

Since $A$ is the orthogonal projection of $Q$ onto the line $PM=MP$, that is, onto the line spanned by $\mathbf p$,

$$A=\frac{\mathbf p\cdot\mathbf q}{|\mathbf p|^2}\mathbf p.$$

Similarly,

$$B=\frac{\mathbf p\cdot\mathbf q}{|\mathbf q|^2}\mathbf q.$$

Then

$$AB^2 =(\mathbf p\cdot\mathbf q)^2 \left| \frac{\mathbf p}{|\mathbf p|^2} - \frac{\mathbf q}{|\mathbf q|^2} \right|^2.$$

Expanding,

$$\left| \frac{\mathbf p}{|\mathbf p|^2} - \frac{\mathbf q}{|\mathbf q|^2} \right|^2 = \frac1{|\mathbf p|^2} +\frac1{|\mathbf q|^2} -\frac{2,\mathbf p\cdot\mathbf q}{|\mathbf p|^2|\mathbf q|^2}.$$

Multiplying by $(\mathbf p\cdot\mathbf q)^2$ gives

$$AB^2 = \frac{(\mathbf p\cdot\mathbf q)^2\bigl(|\mathbf p|^2+|\mathbf q|^2-2\mathbf p\cdot\mathbf q\bigr)} {|\mathbf p|^2|\mathbf q|^2}.$$

Since

$$|\mathbf p|^2+|\mathbf q|^2-2\mathbf p\cdot\mathbf q = |\mathbf p-\mathbf q|^2 = PQ^2,$$

and

$$\frac{\mathbf p\cdot\mathbf q}{|\mathbf p||\mathbf q|} = \cos\angle PMQ,$$

we obtain

$$AB=PQ,|\cos\angle PMQ|.$$

This is remarkably simple. The problem is now to choose $M\in l$ so that $|\cos\angle PMQ|$ is minimal.

The smallest possible value of $|\cos\theta|$ is $0$, attained when $\theta=90^\circ$. Hence we must determine whether there exists a point $M$ on $l$ such that

$$\angle PMQ=90^\circ.$$

By the theorem of Thales, the locus of points $M$ satisfying $\angle PMQ=90^\circ$ is the circle with diameter $PQ$. Since $P$ and $Q$ lie on the same side of $l$, the segment $PQ$ lies entirely on that side, so the midpoint of $PQ$ is on that side. The circle with diameter $PQ$ has radius $\frac12PQ$, which is larger than the distance from its center to $l$ because that distance does not exceed the distance from the center to either endpoint. Hence the circle intersects $l$ in two points. Thus such an $M$ always exists, and then $AB=0$.

The only delicate point is proving rigorously that the circle with diameter $PQ$ must meet $l$.

Problem Understanding

We are given a line $l$ and two points $P,Q$ lying on the same side of it. For a point $M\in l$, let $A$ and $B$ be the feet of the altitudes of triangle $PQM$ dropped onto $PM$ and $QM$ respectively. We must choose $M$ so that the distance $AB$ is as small as possible.

This is a Type C problem.

The answer should be that the minimum value is $0$. Intuitively, if $\angle PMQ=90^\circ$, then the altitude from $Q$ to $PM$ ends at $M$, and the altitude from $P$ to $QM$ also ends at $M$, so the two feet coincide.

The core difficulty is to express $AB$ in a useful form and relate it to a geometric condition on $M$.

Proof Architecture

First, express the feet of the altitudes as orthogonal projections and derive the identity

$$AB=PQ,|\cos\angle PMQ|.$$

This follows from a vector computation.

Second, show that the minimum possible value of $AB$ is at least $0$, and that $AB=0$ exactly when $\angle PMQ=90^\circ$.

Third, prove that there exists a point $M\in l$ with $\angle PMQ=90^\circ$.

The existence follows because the locus $\angle PMQ=90^\circ$ is the circle with diameter $PQ$, and this circle intersects $l$.

The most delicate step is the derivation of the formula

$$AB=PQ,|\cos\angle PMQ|.$$

Solution

Let $A$ be the foot of the altitude from $Q$ onto $PM$, and let $B$ be the foot of the altitude from $P$ onto $QM$.

Choose coordinates with origin at $M$. Let

$$P=\mathbf p,\qquad Q=\mathbf q.$$

Since $A$ is the orthogonal projection of $Q$ onto the line through $M$ and $P$, we have

$$A=\frac{\mathbf p\cdot\mathbf q}{|\mathbf p|^2}\mathbf p.$$

Similarly,

$$B=\frac{\mathbf p\cdot\mathbf q}{|\mathbf q|^2}\mathbf q.$$

Hence

$$AB^2 = \left| \frac{\mathbf p\cdot\mathbf q}{|\mathbf p|^2}\mathbf p - \frac{\mathbf p\cdot\mathbf q}{|\mathbf q|^2}\mathbf q \right|^2.$$

Factoring out $(\mathbf p\cdot\mathbf q)^2$,

$$AB^2 = (\mathbf p\cdot\mathbf q)^2 \left| \frac{\mathbf p}{|\mathbf p|^2} - \frac{\mathbf q}{|\mathbf q|^2} \right|^2.$$

Expanding the square,

$$AB^2 = (\mathbf p\cdot\mathbf q)^2 \left( \frac1{|\mathbf p|^2} + \frac1{|\mathbf q|^2} - \frac{2,\mathbf p\cdot\mathbf q}{|\mathbf p|^2|\mathbf q|^2} \right).$$

Therefore

$$AB^2 = \frac{(\mathbf p\cdot\mathbf q)^2 \bigl(|\mathbf p|^2+|\mathbf q|^2-2\mathbf p\cdot\mathbf q\bigr)} {|\mathbf p|^2|\mathbf q|^2}.$$

Since

$$|\mathbf p|^2+|\mathbf q|^2-2\mathbf p\cdot\mathbf q = |\mathbf p-\mathbf q|^2 = PQ^2,$$

we obtain

$$AB^2 = PQ^2 \left( \frac{\mathbf p\cdot\mathbf q}{|\mathbf p||\mathbf q|} \right)^2.$$

The angle between the vectors $\mathbf p$ and $\mathbf q$ is $\angle PMQ$, hence

$$\frac{\mathbf p\cdot\mathbf q}{|\mathbf p||\mathbf q|} = \cos\angle PMQ.$$

Thus

$$AB=PQ,|\cos\angle PMQ|.$$

Since $PQ$ is fixed, minimizing $AB$ is equivalent to minimizing $|\cos\angle PMQ|$. The least possible value of $|\cos\angle PMQ|$ is $0$, attained precisely when

$$\angle PMQ=90^\circ.$$

It remains to show that such a point $M$ exists on the line $l$.

The locus of points $X$ satisfying

$$\angle PXQ=90^\circ$$

is the circle with diameter $PQ$.

Let $O$ be the midpoint of $PQ$. Because $P$ and $Q$ lie on the same side of $l$, the segment $PQ$ lies entirely on that side, so $O$ also lies on that side of $l$.

Let $d$ be the distance from $O$ to $l$. Since $O$ lies on the segment $PQ$,

$$d\le OP=\frac{PQ}{2}.$$

Equality would imply that one endpoint of the segment $PQ$ lies on $l$, contrary to the assumption that both $P$ and $Q$ are on the same side of $l$. Hence

$$d<\frac{PQ}{2}.$$

The circle with center $O$ and radius $\frac{PQ}{2}$ is exactly the circle with diameter $PQ$. Since the distance from its center to $l$ is smaller than its radius, the line $l$ intersects this circle in two points. For either intersection point $M$,

$$\angle PMQ=90^\circ.$$

For such a point,

$$AB=PQ,|\cos90^\circ|=0.$$

Since distances are nonnegative, no smaller value is possible. Thus the minimum value is

$$\boxed{0},$$

and it is attained at either intersection of $l$ with the circle having diameter $PQ$.

Verification of Key Steps

The projection formula must be checked carefully. For a line spanned by a nonzero vector $\mathbf p$, the orthogonal projection of $\mathbf q$ onto that line is

$$\frac{\mathbf p\cdot\mathbf q}{|\mathbf p|^2}\mathbf p.$$

Substituting this expression for both altitude feet and expanding produces

$$AB^2 = \frac{(\mathbf p\cdot\mathbf q)^2|\mathbf p-\mathbf q|^2} {|\mathbf p|^2|\mathbf q|^2},$$

which simplifies to

$$AB^2=PQ^2\cos^2\angle PMQ.$$

No geometric assumptions beyond the definitions are used.

The existence of an intersection between $l$ and the circle with diameter $PQ$ depends on the inequality

$$d<\frac{PQ}{2}.$$

Here $d$ is the distance from the midpoint $O$ of $PQ$ to $l$. Since $O$ lies on the segment joining two points strictly on the same side of $l$, the whole segment lies in the corresponding open half-plane. Consequently $O$ is not on $l$, and the perpendicular distance from $O$ to $l$ is strictly smaller than the distance from $O$ to either endpoint of the segment, namely $\frac{PQ}{2}$. This guarantees intersection.

When $\angle PMQ=90^\circ$, both altitudes indeed end at $M$. The side $PM$ is perpendicular to $QM$, so the altitude from $Q$ to $PM$ is the segment $QM$, whose foot is $M$. Similarly the altitude from $P$ to $QM$ has foot $M$. Hence $A=B=M$, giving distance $0$.

Alternative Approaches

A synthetic solution can avoid vectors. Let $\theta=\angle PMQ$. In triangle $PQM$, the feet $A$ and $B$ satisfy

$$MA=MQ\cos\theta,\qquad MB=MP\cos\theta.$$

Since $A$ and $B$ lie on the rays $MP$ and $MQ$ respectively, the law of cosines in triangle $AMB$ yields

$$AB^2 = \cos^2\theta, \bigl(MP^2+MQ^2-2,MP\cdot MQ\cos\theta\bigr).$$

The expression in parentheses equals $PQ^2$ by the law of cosines in triangle $PQM$, giving

$$AB=PQ,|\cos\theta|.$$

The remainder is identical: the minimum is attained when $\theta=90^\circ$, and such points are the intersections of $l$ with the circle having diameter $PQ$.

The vector method is preferable because the identity for $AB$ emerges in a single computation and immediately reduces the problem to a classical locus theorem.