Kvant Math Problem 87

Let the three circles have common radius $r$, and let their common point be $P$.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m04s
Source on kvant.digital

Problem

Prove that if three circles of equal radius pass through a single point, then the three other points of pairwise intersection of these circles lie on a circle of the same radius (Fig. 1).

Insert illustration

Exploration

Let the three circles have common radius $r$, and let their common point be $P$. Denote the circles by $\omega_1,\omega_2,\omega_3$. Let the second intersection points of the pairs $(\omega_2,\omega_3)$, $(\omega_3,\omega_1)$, $(\omega_1,\omega_2)$ be $A,B,C$ respectively.

The statement claims that $A,B,C$ lie on a circle of radius $r$.

A direct coordinate computation looks possible. Place $P$ at the origin. Since each circle passes through $P$ and has radius $r$, its center lies at distance $r$ from $P$. If the center of $\omega_i$ is the vector $u_i$ with $|u_i|=r$, then the equation of $\omega_i$ is

$$x^2+y^2-2u_i\cdot (x,y)=0.$$

Subtracting the equations for two circles gives a line. Solving simultaneously suggests that the second intersection point of $\omega_i$ and $\omega_j$ should be $u_i+u_j$.

This deserves checking carefully. Take

$$|x|^2-2u_i\cdot x=0,\qquad |x|^2-2u_j\cdot x=0.$$

Subtracting yields

$$(u_i-u_j)\cdot x=0.$$

Now test $x=u_i+u_j$:

$$|u_i+u_j|^2-2u_i\cdot(u_i+u_j) =|u_j|^2-|u_i|^2=0.$$

Thus $u_i+u_j$ indeed lies on both circles. Since $0$ is one intersection point, the other is $u_i+u_j$.

Then

$$A=u_2+u_3,\qquad B=u_3+u_1,\qquad C=u_1+u_2.$$

The next question is whether these three points lie on a circle of radius $r$. Their centroid is

$$\frac{A+B+C}{2}=u_1+u_2+u_3.$$

Compute distances from this point:

$$|A-(u_1+u_2+u_3)|=|{-u_1}|=r,$$

and similarly for $B,C$.

This is extremely short. The only potentially dangerous point is the identification of the second intersection point as $u_i+u_j$. One must prove that this point is distinct from $0$ unless the circles coincide. If two centers coincide, then the two circles are identical and the problem degenerates. The intended problem clearly assumes three distinct circles, so each pair has a genuine second intersection point.

The core insight is that, in vector form, the second intersection point of two equal circles through the origin is the sum of their center vectors.

Problem Understanding

We are given three distinct circles of the same radius, all passing through a common point $P$. Each pair of circles intersects at one additional point besides $P$. Denote these three additional intersection points by $A,B,C$. The task is to prove that $A,B,C$ lie on a circle whose radius equals the common radius of the original circles.

This is a Type B problem. The statement itself must be proved.

The core difficulty is identifying a useful relation between the centers of the original circles and the new intersection points. Once the correct vector description is found, the conclusion becomes a direct computation.

Proof Architecture

The proof places the common point $P$ at the origin of the plane and represents the centers of the circles by vectors $u_1,u_2,u_3$ of equal length $r$, because every center is at distance $r$ from $P$.

The first lemma states that the circle with center $u_i$ and radius $r$ has equation

$$|x|^2-2u_i\cdot x=0.$$

This follows from expanding the standard equation of a circle.

The second lemma states that for $i\ne j$, the second intersection point of the circles centered at $u_i$ and $u_j$ is the vector $u_i+u_j$. This is proved by direct substitution into both circle equations.

The third lemma states that the points

$$A=u_2+u_3,\qquad B=u_3+u_1,\qquad C=u_1+u_2$$

all lie on the circle centered at

$$S=u_1+u_2+u_3$$

with radius $r$. This follows by computing the distances from $S$.

The second lemma is the point most likely to fail under insufficient justification, because one must verify carefully that the proposed point actually lies on both circles.

Solution

Let the common point of the three circles be $P$. Choose coordinates so that $P=(0,0)$.

Let $\omega_i$ have center $u_i$ for $i=1,2,3$. Since all three circles have radius $r$ and pass through the origin,

$$|u_1|=|u_2|=|u_3|=r.$$

A point $x$ lies on $\omega_i$ precisely when

$$|x-u_i|=r.$$

Squaring and using $|u_i|=r$ gives

$$|x|^2-2u_i\cdot x+|u_i|^2=r^2,$$

hence

$$|x|^2-2u_i\cdot x=0.$$

Let $A$ be the second intersection point of $\omega_2$ and $\omega_3$. We claim that

$$A=u_2+u_3.$$

Indeed,

\begin{align*}

|u_2+u_3|^2-2u_2\cdot(u_2+u_3)

&=(|u_2|^2+2u_2\cdot u_3+|u_3|^2)-2|u_2|^2-2u_2\cdot u_3 \

&=|u_3|^2-|u_2|^2 \

&=0.

\end{align*}

Thus $u_2+u_3$ lies on $\omega_2$.

Similarly,

\begin{align*}

|u_2+u_3|^2-2u_3\cdot(u_2+u_3)

&=|u_2|^2-|u_3|^2 \

&=0,

\end{align*}

so $u_2+u_3$ also lies on $\omega_3$.

Therefore $u_2+u_3$ is an intersection point of $\omega_2$ and $\omega_3$. Since the other intersection point is the origin, we obtain

$$A=u_2+u_3.$$

By the same argument,

$$B=u_3+u_1,\qquad C=u_1+u_2.$$

Now define

$$S=u_1+u_2+u_3.$$

Then

$$A-S=-u_1,$$

so

$$|A-S|=|u_1|=r.$$

Likewise,

$$B-S=-u_2,\qquad C-S=-u_3,$$

hence

$$|B-S|=|C-S|=r.$$

Thus the three points $A,B,C$ all lie on the circle centered at $S$ with radius $r$. This radius equals the common radius of the original circles.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the derivation of the circle equation

$$|x|^2-2u_i\cdot x=0.$$

Starting from

$$|x-u_i|=r,$$

one gets

$$|x-u_i|^2=r^2.$$

Expanding gives

$$|x|^2-2u_i\cdot x+|u_i|^2=r^2.$$

Since the origin lies on the circle, the distance from the center to the origin equals the radius, so $|u_i|=r$. Substituting this identity yields the equation above. Forgetting to use $|u_i|=r$ would leave an incorrect constant term.

The second delicate step is proving that $u_i+u_j$ lies on both circles. A careless argument might use symmetry without computation. The substitution must be carried out explicitly:

\begin{align*}

|u_i+u_j|^2-2u_i\cdot(u_i+u_j)

&=|u_j|^2-|u_i|^2.

\end{align*}

The equality vanishes only because the radii are equal. If the circles had different radii, this cancellation would fail.

The third delicate step is identifying the center of the final circle. The point

$$S=u_1+u_2+u_3$$

is not guessed arbitrarily. It is forced by the identities

$$A=u_2+u_3,\qquad B=u_3+u_1,\qquad C=u_1+u_2.$$

Subtracting gives

$$A-S=-u_1,\qquad B-S=-u_2,\qquad C-S=-u_3.$$

All three distances equal $r$ because the vectors $u_i$ all have length $r$.

Alternative Approaches

A synthetic proof can be obtained using rotations. Let $O_i$ be the center of $\omega_i$. Since $PO_iA O_j$ is a rhombus for the appropriate indices, the segments joining centers and new intersection points are parallel and equal. One can then express the sides of the triangle formed by the new intersection points in terms of the triangle $O_1O_2O_3$. After that, a direct geometric construction produces the desired circle.

Another approach uses complex numbers. Place the common point at the origin and represent the centers by complex numbers $u_1,u_2,u_3$ with equal modulus $r$. The equation of each circle becomes

$$z\bar z=u_i\bar z+\bar u_i z.$$

The second intersection point of the circles corresponding to $u_i$ and $u_j$ is then $u_i+u_j$. The rest of the proof is identical to the vector argument.

The vector method is preferable because every computation is linear after the initial setup, and the center of the final circle emerges immediately from the coordinates of the three intersection points.