Kvant Math Problem 1511

Let the two circles be $\Gamma_1$ and $\Gamma_2$, with centers $O_1$ and $O_2$.

kvantmathematicsolympiad

Verified: yes
Verdicts: PASS + PASS
Solve time: 43m02s
Source on kvant.digital

Problem

On the plane, two intersecting circles are given. Point $A$ is one of the two points of intersection of these circles. In each circle, a diameter is drawn parallel to the tangent to the other circle at point $A$, and these diameters do not intersect each other. Prove that the endpoints of these diameters lie on a single circle.

S. Berlov

Saint Petersburg City Mathematical Olympiad (1995)

Exploration

Let the two circles be $\Gamma_1$ and $\Gamma_2$, with centers $O_1$ and $O_2$. They intersect at the point $A$.

Let $CD$ be the diameter of $\Gamma_1$ parallel to the tangent to $\Gamma_2$ at $A$, and let $EF$ be the diameter of $\Gamma_2$ parallel to the tangent to $\Gamma_1$ at $A$.

Choose Cartesian coordinates with origin at $A$.

Write

$$\mathbf u=\overrightarrow{AO_1},\qquad \mathbf v=\overrightarrow{AO_2},$$

and let

$$R_1=|\mathbf u|,\qquad R_2=|\mathbf v|.$$

Let $J$ denote rotation by $90^\circ$.

Since the tangent to $\Gamma_2$ at $A$ is perpendicular to $AO_2$, the diameter $CD$ is parallel to $J\mathbf v$. Likewise, $EF$ is parallel to $J\mathbf u$.

Problem Understanding

The endpoints of the two diameters are

$$C,D\in\Gamma_1,\qquad E,F\in\Gamma_2.$$

The goal is to prove that the four points $C,D,E,F$ lie on a common circle.

The strategy is to write these four points explicitly in vector form and then show that they satisfy a single quadratic equation of the form

$$|\mathbf x|^2-2\mathbf c\cdot\mathbf x+d=0,$$

which is the equation of a circle.

Proof Architecture

First determine the coordinates of the four endpoints.

Then introduce a quadratic expression

$$Q(\mathbf x) = |\mathbf x|^2-2(\mathbf u+\mathbf v)\cdot\mathbf x+(R_1^2+R_2^2).$$

A direct computation will show that $Q$ takes the same value at $C,D,E,F$.

That common value will then be absorbed into the constant term, producing a circle equation satisfied by all four points.

Solution

Since $CD$ is a diameter of $\Gamma_1$ through the center $O_1$ and is parallel to $J\mathbf v$, its endpoints have the form

$$C=\mathbf u+t,J\mathbf v, \qquad D=\mathbf u-t,J\mathbf v.$$

Because $C$ lies on $\Gamma_1$,

$$|C-\mathbf u|=R_1.$$

Since $|J\mathbf v|=R_2$, it follows that

$$tR_2=R_1,$$

hence

$$t=\frac{R_1}{R_2}.$$

Thus

$$C=\mathbf u+\frac{R_1}{R_2}J\mathbf v, \qquad D=\mathbf u-\frac{R_1}{R_2}J\mathbf v.$$

Similarly,

$$E=\mathbf v+\frac{R_2}{R_1}J\mathbf u, \qquad F=\mathbf v-\frac{R_2}{R_1}J\mathbf u.$$

Let

$$k=\frac{R_1}{R_2}.$$

Then

$$C=\mathbf u+kJ\mathbf v, \qquad D=\mathbf u-kJ\mathbf v,$$

and

$$E=\mathbf v+\frac1kJ\mathbf u, \qquad F=\mathbf v-\frac1kJ\mathbf u.$$

Consider

$$Q(\mathbf x) = |\mathbf x|^2 -2(\mathbf u+\mathbf v)\cdot\mathbf x +(R_1^2+R_2^2).$$

For $C$,

$$|C|^2 = |\mathbf u+kJ\mathbf v|^2 = R_1^2+k^2R_2^2 +2k,\mathbf u\cdot J\mathbf v.$$

Since $k^2R_2^2=R_1^2$,

$$|C|^2 = 2R_1^2+2k,\mathbf u\cdot J\mathbf v.$$

Also,

$$(\mathbf u+\mathbf v)\cdot C = R_1^2+\mathbf u\cdot\mathbf v +k,\mathbf u\cdot J\mathbf v.$$

Substituting into $Q$ gives

$$Q(C) = R_1^2+R_2^2 -2,\mathbf u\cdot\mathbf v.$$

Replacing $k$ by $-k$ yields the same value for $D$:

$$Q(D) = R_1^2+R_2^2 -2,\mathbf u\cdot\mathbf v.$$

For $E$,

$$|E|^2 = \left|\mathbf v+\frac1kJ\mathbf u\right|^2 = R_2^2+\frac1{k^2}R_1^2 +\frac2k,\mathbf v\cdot J\mathbf u.$$

Since $\frac1{k^2}R_1^2=R_2^2$,

$$|E|^2 = 2R_2^2+\frac2k,\mathbf v\cdot J\mathbf u.$$

Moreover,

$$(\mathbf u+\mathbf v)\cdot E = \mathbf u\cdot\mathbf v +R_2^2 +\frac1k,\mathbf v\cdot J\mathbf u.$$

Hence

$$Q(E) = R_1^2+R_2^2 -2,\mathbf u\cdot\mathbf v.$$

The same computation gives

$$Q(F) = R_1^2+R_2^2 -2,\mathbf u\cdot\mathbf v.$$

Thus

$$Q(C)=Q(D)=Q(E)=Q(F) = R_1^2+R_2^2-2,\mathbf u\cdot\mathbf v.$$

Since

$$R_1^2+R_2^2-2,\mathbf u\cdot\mathbf v = |\mathbf u-\mathbf v|^2,$$

all four points satisfy

$$Q(\mathbf x)=|\mathbf u-\mathbf v|^2.$$

Substituting the definition of $Q$ and moving the common value to the left side gives

$$|\mathbf x|^2 -2(\mathbf u+\mathbf v)\cdot\mathbf x +(R_1^2+R_2^2) -|\mathbf u-\mathbf v|^2 =0.$$

Because

$$(R_1^2+R_2^2)-|\mathbf u-\mathbf v|^2 = 2,\mathbf u\cdot\mathbf v,$$

this equation becomes

$$|\mathbf x|^2 -2(\mathbf u+\mathbf v)\cdot\mathbf x +2,\mathbf u\cdot\mathbf v =0.$$

This is an equation of a circle. Every one of the points $C,D,E,F$ satisfies it, so the four endpoints lie on a single circle.

Verification of Key Steps

The only correction required to the previous solution concerns the final stage. After computing

$$Q(C)=Q(D)=Q(E)=Q(F),$$

there is no need to prove that this common value is zero. In fact,

$$Q(C)=Q(D)=Q(E)=Q(F) = |\mathbf u-\mathbf v|^2,$$

which is generally positive because the centers are distinct.

The common value itself is sufficient. Subtracting it from $Q$ produces a circle equation satisfied by all four points.

Alternative Approaches

A synthetic proof can be obtained by expressing the relevant angles through the tangent-chord theorem and showing that a pair of opposite angles determined by the four endpoints sum to $180^\circ$. This also yields the concyclicity of $C,D,E,F$.