Kvant Math Problem 630
The point $M$ is defined from the circle through $P,Q,K$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 22m44s
Source on kvant.digital
Problem
In the plane, a circle and points $P$, $Q$ on it are given. Draw a circle through the points $P$, $Q$ and an arbitrary point $K$ of the plane. Let $M$ be the point of intersection of the tangent to this circle at the point $K$ with the line $PQ$. What set is traced out by the points $M$?
I. F. Sharygin
Exploration
The point $M$ is defined from the circle through $P,Q,K$. Since $P$ and $Q$ are fixed, all such circles form the pencil of circles through $P$ and $Q$. The tangent at $K$ to the chosen circle meets the fixed line $PQ$ at $M$.
The first task is to express $M$ in terms of $K$. Let the variable circle be $\omega$. Since $M$ lies on the tangent at $K$, the power of $M$ with respect to $\omega$ equals $MK^{2}$. Since $P,Q\in\omega$ and $M\in PQ$, the secant theorem gives
$$MP\cdot MQ=MK^{2}.$$
This relation involves only $M$ and $K$, not the particular circle. It is natural to ask whether it already characterizes the construction.
Choose coordinates on the line $PQ$. Let the midpoint of $PQ$ be the origin and let
$$P=(-a,0),\qquad Q=(a,0).$$
If $M=(t,0)$, then
$$MP\cdot MQ=(t+a)(t-a)=t^{2}-a^{2}.$$
Hence
$$MK^{2}=t^{2}-a^{2}.$$
Now let $K=(x,y)$. Since
$$MK^{2}=(x-t)^{2}+y^{2},$$
we obtain
$$(x-t)^{2}+y^{2}=t^{2}-a^{2}.$$
After expansion,
$$x^{2}+y^{2}-2xt+a^{2}=0.$$
For $x\neq0$,
$$t=\frac{x^{2}+y^{2}+a^{2}}{2x}.$$
This formula suggests using the circle
$$\Gamma:\ x^{2}+y^{2}=a^{2},$$
the given circle. Indeed, for a point $K=(x,y)$, the polar of $K$ with respect to $\Gamma$ is
$$xx_{0}+yy_{0}=a^{2},$$
that is,
$$xx'+yy'=a^{2}.$$
Its intersection with the $x$-axis is
$$\left(\frac{a^{2}}{x},0\right).$$
Our point $M$ is not that intersection. However,
$$t=\frac12\left(x+\frac{a^{2}+y^{2}}{x}\right),$$
which hints at harmonic or pole-polar geometry.
A more direct route is better. Rewrite
$$2xt=x^{2}+y^{2}+a^{2}.$$
This is exactly the equation of the circle with diameter joining the origin and $M=(t,0)$:
$$x^{2}+y^{2}-2tx=0.$$
The extra term $a^{2}$ shows
$$x^{2}+y^{2}-2tx=-a^{2}.$$
Hence for fixed $M$, the locus of all corresponding points $K$ is a circle centered at $(t,0)$ with radius $\sqrt{t^{2}-a^{2}}$.
Such a circle exists iff $t^{2}-a^{2}\ge0$, that is, $|t|\ge a$. Thus every point $M$ on the line $PQ$ outside the segment $PQ$ arises, while no interior point does.
The crucial point is proving the converse: for every point $M$ with $|t|>a$, there exists at least one point $K$ and a circle through $P,Q,K$ whose tangent at $K$ passes through $M$.
Problem Understanding
We are given a fixed circle and two fixed points $P,Q$ on it. For an arbitrary point $K$ of the plane, we draw the unique circle through $P,Q,K$ and let $M$ be the intersection of the tangent at $K$ to this circle with the line $PQ$. The problem asks for the locus of all possible points $M$.
This is a Type A problem. We must determine exactly which points on the line $PQ$ occur and prove that no others occur.
The expected answer is that the locus consists of the two rays of the line $PQ$ lying outside the segment $PQ$, including the endpoints $P$ and $Q$. The relation
$$MK^{2}=MP\cdot MQ$$
suggests this because the left-hand side is nonnegative, forcing $MP\cdot MQ\ge0$, which means that $M$ cannot lie inside the segment $PQ$.
Proof Architecture
The first lemma is that for every admissible configuration,
$$MK^{2}=MP\cdot MQ.$$
This follows from the power of the point $M$ with respect to the circle through $P,Q,K$.
The second lemma is that every resulting point $M$ lies on the line $PQ$ outside the open segment $PQ$.
This follows because $MK^{2}\ge0$.
The third lemma is that for any point $M$ on the line $PQ$ with $MP\cdot MQ\ge0$, there exists a circle through $P,Q$ whose power at $M$ equals $MP\cdot MQ$ and whose tangent from $M$ touches it at some point $K$.
This constructs a corresponding configuration.
The hardest direction is the converse existence statement. The most delicate point is showing that every exterior point of the line $PQ$ actually occurs.
Solution
Let $\omega$ be the circle through $P,Q,K$, and let the tangent to $\omega$ at $K$ meet the line $PQ$ at $M$.
Since $M$ lies on the tangent at $K$, the power of $M$ with respect to $\omega$ is
$$\operatorname{Pow}_{\omega}(M)=MK^{2}.$$
On the other hand, the line $PQ$ is a secant of $\omega$ through $M$, meeting $\omega$ at $P$ and $Q$. By the secant theorem,
$$\operatorname{Pow}_{\omega}(M)=MP\cdot MQ.$$
Hence
$$MK^{2}=MP\cdot MQ.$$
Since $MK^{2}\ge0$, we obtain
$$MP\cdot MQ\ge0.$$
Along the line $PQ$, the product $MP\cdot MQ$ is negative exactly for points strictly between $P$ and $Q$, zero at $P$ and $Q$, and positive outside the segment $PQ$. Therefore every point arising from the construction belongs to the union of the two closed rays of the line $PQ$ issuing from $P$ and $Q$ away from the segment $PQ$.
It remains to prove the converse.
Take an arbitrary point $M$ on the line $PQ$ such that
$$MP\cdot MQ\ge0.$$
If $M=P$ or $M=Q$, choose $K=P$ or $K=Q$. Then the tangent at that point meets the line $PQ$ at the same point, so the endpoints of the claimed locus are obtained.
Now assume $MP\cdot MQ>0$. Let
$$r=\sqrt{MP\cdot MQ}.$$
Choose any point $K$ satisfying
$$MK=r.$$
Such points form the circle centered at $M$ of radius $r$.
Consider the unique circle $\omega$ through $P,Q,K$. Since $M$ lies on the secant line $PQ$, its power with respect to $\omega$ is
$$\operatorname{Pow}_{\omega}(M)=MP\cdot MQ=r^{2}.$$
Since $MK=r$, this becomes
$$MK^{2}=\operatorname{Pow}_{\omega}(M).$$
A standard converse of the tangent-power theorem states that if a point $X$ of a circle $\omega$ satisfies
$$MX^{2}=\operatorname{Pow}_{\omega}(M),$$
then $MX$ is tangent to $\omega$ at $X$. For completeness, let $T$ be the second intersection of the line $MX$ with $\omega$. The secant theorem gives
$$MX\cdot MT=\operatorname{Pow}_{\omega}(M).$$
Using $MX^{2}=\operatorname{Pow}_{\omega}(M)$, we obtain
$$MX\cdot MT=MX^{2}.$$
Since $MX\neq0$, it follows that
$$MT=MX.$$
The points $M,X,T$ lie on the same line, and the only point of the line $MX$ on the circle at distance $MX$ from $M$ is $X$ itself. Hence $T=X$. Thus the line $MX$ meets $\omega$ with multiplicity two at $X$, which means that $MX$ is tangent to $\omega$ at $X$.
Applying this with $X=K$, the line $MK$ is tangent to $\omega$ at $K$. By construction $M\in PQ$, so the tangent at $K$ meets the line $PQ$ exactly at $M$.
Thus every point $M$ with $MP\cdot MQ\ge0$ is obtained from the construction.
We have proved both inclusions. Hence the locus of points $M$ is precisely the set of points on the line $PQ$ lying outside the open segment $PQ$, together with the endpoints $P$ and $Q$.
$$\boxed{\text{The locus is the union of the two closed rays of the line }PQ\text{ exterior to the segment }PQ.}$$
Verification of Key Steps
The identity
$$MK^{2}=MP\cdot MQ$$
comes from two independent expressions for the power of $M$ with respect to the circle through $P,Q,K$. A common mistake is to write $MK=MP\cdot MQ$ instead of the squared relation. The tangent theorem always gives the square of the tangent length.
To determine the sign of $MP\cdot MQ$, one must examine the position of $M$ on the line $PQ$. If $M$ lies strictly between $P$ and $Q$, the directed distances to $P$ and $Q$ have opposite signs, so the product is negative. Outside the segment they have the same sign, so the product is positive.
For the converse, the key point is not to construct an arbitrary circle centered at $M$. Instead, one first chooses a point $K$ with
$$MK=\sqrt{MP\cdot MQ},$$
then takes the unique circle through $P,Q,K$. Its power at $M$ is automatically $MP\cdot MQ$, because $P$ and $Q$ lie on the circle and $M$ lies on the secant $PQ$. The equality
$$MK^{2}=\operatorname{Pow}_{\omega}(M)$$
then forces $MK$ to be tangent to $\omega$ at $K$, giving the required configuration.
Alternative Approaches
A projective-geometric approach uses pole-polar duality with respect to the fixed circle. Let $M$ be the intersection of the tangent at $K$ with the chord line $PQ$. By a standard theorem for a complete quadrilateral inscribed in a circle, the point $K$ lies on the polar of $M$ with respect to the fixed circle through $P$ and $Q$. The condition that the polar be real is equivalent to $M$ lying outside or on the fixed circle, and because $M$ is constrained to the line $PQ$, this means precisely that $M$ lies outside the segment $PQ$.
The power-of-a-point argument is preferable because it requires only the tangent-secant theorem and yields the locus immediately from the single identity
$$MK^{2}=MP\cdot MQ.$$
It also provides the converse construction in a completely elementary way.