Kvant Math Problem 611
The statement involves two circles.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m36s
Source on kvant.digital
Problem
An arbitrary point $M$ is chosen on the chord $AB$ of a circle with center $O$. A circle is drawn through the points $A$, $M$, and $O$; it intersects the first circle at the points $A$ and $C$. Prove that $|MB|=|MC|$.
S. Kolpakov, 10th-grade student
Exploration
The statement involves two circles. The first circle has center $O$, and $AB$ is one of its chords. A point $M$ lies on $AB$. The second circle passes through $A$, $M$, and $O$, and meets the first circle again at $C$.
The conclusion $MB=MC$ suggests that $M$ lies on the perpendicular bisector of $BC$, or equivalently that some angle relation forces $M$ to be the center of a circle through $B$ and $C$. Since $B$ and $C$ lie on the first circle, it is natural to search for an angle equality involving $\angle MBC$ and $\angle BCM$.
Let the first circle be denoted by $\omega$ and the second by $\Gamma$.
Because $A,M,B$ are collinear, the angle $\angle OMA$ equals $\angle OMB$. Since $A,O,M,C$ lie on $\Gamma$, opposite angles satisfy
$$\angle OMA+\angle OCA=180^\circ.$$
Hence
$$\angle OMB+\angle OCA=180^\circ.$$
Since $O,B,C,A$ lie on $\omega$, the quadrilateral $OABC$ is cyclic. Therefore
$$\angle OCA+\angle OBA=180^\circ.$$
Comparing the two equalities gives
$$\angle OMB=\angle OBA.$$
Because $M$ lies on $BA$, this means
$$\angle OMB=\angle OBM.$$
Thus triangle $OMB$ is isosceles and
$$OM=OB.$$
Since $OB$ and $OC$ are radii of the first circle, we obtain
$$OM=OB=OC.$$
Hence $O,B,M,C$ lie on a circle centered at $O$. Then equal angles subtending the same chord may produce the desired result. Indeed,
$$\angle MBC=\angle MOC,$$
and because $A,O,M,C$ are cyclic,
$$\angle MOC=\angle MAC.$$
Since $A,M,B$ are collinear,
$$\angle MAC=\angle BAC.$$
Finally, on $\omega$,
$$\angle BAC=\angle BOC.$$
Thus $\angle MBC=\angle BOC$.
But in the circle through $O,B,M,C$, the angle $\angle BOC$ subtends chord $BC$, so
$$\angle BOC=\angle BMC.$$
This does not directly give an isosceles triangle. A cleaner route is needed.
From the circle through $O,B,M,C$,
$$\angle MCB=\angle MOB.$$
Since $A,O,M,C$ are cyclic,
$$\angle MOB=\angle MOA.$$
Again in that circle,
$$\angle MOA=\angle MCA.$$
Because $A,M,B$ are collinear,
$$\angle MCA=\angle MCB.$$
This chain is circular and does not produce new information.
A more systematic approach is to use the circle $O,B,M,C$. In that circle,
$$\angle MCB=\angle MOB.$$
Since $M$ lies on $AB$,
$$\angle MOB=\angle MOA.$$
In $\Gamma$,
$$\angle MOA=\angle MCA.$$
Because $A,B,M$ are collinear,
$$\angle MCA=\angle MCB.$$
This confirms consistency but not the claim.
A better target is to prove
$$\angle MCB=\angle MBC.$$
From the circle $O,B,M,C$,
$$\angle MCB=\angle MOB.$$
Since $A,B,M$ are collinear,
$$\angle MOB=\angle MOA.$$
From $\Gamma$,
$$\angle MOA=\angle MCA.$$
Since $A,B,M$ are collinear,
$$\angle MCA=\angle BCA.$$
On $\omega$,
$$\angle BCA=\angle BOA.$$
Because $M$ lies on $AB$,
$$\angle BOA=\angle BOM.$$
Again in the circle $O,B,M,C$,
$$\angle BOM=\angle BCM.$$
Hence $\angle MCB=\angle BCM$, which yields $MB=MC$.
The delicate step is proving $OM=OB$, which creates the auxiliary circle through $O,B,M,C$.
Problem Understanding
We are given a circle $\omega$ with center $O$ and a chord $AB$. A point $M$ is chosen on the chord. A second circle $\Gamma$ is drawn through $A$, $M$, and $O$. Its second intersection with $\omega$ is $C$. We must prove that the distances from $M$ to $B$ and from $M$ to $C$ are equal.
This is a Type B problem, a pure proof.
The core difficulty is discovering a non-obvious relation between the two circles. The crucial step is to show that $OM=OB$. Since $OB=OC$ are radii of the original circle, this places $B$ and $C$ on a circle centered at $O$ and passing through $M$, creating a useful cyclic quadrilateral $OBMC$.
Proof Architecture
First, prove that $\angle OMB=\angle OBA$. This follows by comparing opposite-angle relations in the cyclic quadrilaterals $AOMC$ and $AOBC$.
Second, deduce that $OM=OB$. Since $M$ lies on $AB$, the equality $\angle OMB=\angle OBA$ becomes $\angle OMB=\angle OBM$, making triangle $OMB$ isosceles.
Third, deduce that $OM=OB=OC$, hence the points $O,B,M,C$ lie on a circle centered at $O$.
Fourth, prove that $\angle MCB=\angle BCM$ by repeatedly converting angles between the circles $AOMC$, $AOBC$, and $OBMC$.
Once the base angles of triangle $MBC$ are equal, conclude that $MB=MC$.
The lemma most likely to fail under scrutiny is the derivation of $OM=OB$, because it requires careful comparison of two different cyclic quadrilaterals.
Solution
Let $\omega$ be the original circle and let $\Gamma$ be the circle through $A$, $M$, $O$, and $C$.
Since $A,O,M,C$ lie on $\Gamma$, opposite angles of the cyclic quadrilateral $AOMC$ satisfy
$$\angle OMA+\angle OCA=180^\circ.$$
Because $A,M,B$ are collinear,
$$\angle OMA=\angle OMB.$$
Hence
$$\angle OMB+\angle OCA=180^\circ. \tag{1}$$
The points $A,O,B,C$ lie on $\omega$, so $AOBC$ is cyclic. Therefore
$$\angle OBA+\angle OCA=180^\circ. \tag{2}$$
Comparing (1) and (2) yields
$$\angle OMB=\angle OBA.$$
Since $M$ lies on the line $BA$,
$$\angle OBA=\angle OBM.$$
Thus
$$\angle OMB=\angle OBM,$$
and triangle $OMB$ is isosceles. Consequently,
$$OM=OB.$$
Because $OB$ and $OC$ are radii of $\omega$,
$$OB=OC.$$
Therefore
$$OM=OB=OC.$$
Hence the points $O,B,M,C$ lie on the circle with center $O$ and radius $OB$.
Now consider the angle $\angle MCB$. Since $O,B,M,C$ lie on a circle,
$$\angle MCB=\angle MOB. \tag{3}$$
Because $A,B,M$ are collinear,
$$\angle MOB=\angle MOA. \tag{4}$$
Since $A,O,M,C$ lie on $\Gamma$,
$$\angle MOA=\angle MCA. \tag{5}$$
Because $A,B,M$ are collinear,
$$\angle MCA=\angle BCA. \tag{6}$$
Since $A,O,B,C$ lie on $\omega$,
$$\angle BCA=\angle BOA. \tag{7}$$
Again using the collinearity of $A,B,M$,
$$\angle BOA=\angle BOM. \tag{8}$$
Finally, in the circle through $O,B,M,C$,
$$\angle BOM=\angle BCM. \tag{9}$$
Combining (3) through (9), we obtain
$$\angle MCB=\angle BCM.$$
Thus triangle $MBC$ is isosceles, and therefore
$$MB=MC.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the deduction that $OM=OB$. The equality
$$\angle OMB+\angle OCA=180^\circ$$
comes from the cyclic quadrilateral $AOMC$, while
$$\angle OBA+\angle OCA=180^\circ$$
comes from the cyclic quadrilateral $AOBC$. Since both angles are supplementary to the same angle $\angle OCA$, they are equal. Because $A,B,M$ are collinear, $\angle OBA=\angle OBM$. This produces equal base angles in triangle $OMB$, giving $OM=OB$.
The second delicate step is the assertion that $O,B,M,C$ are concyclic. After obtaining $OM=OB$ and using $OB=OC$, all three distances $OM$, $OB$, and $OC$ equal the same radius. Thus $B$, $M$, and $C$ lie on the circle centered at $O$ with radius $OB$. No further cyclicity criterion is needed.
The third delicate step is the chain proving $\angle MCB=\angle BCM$. Each equality comes from a specific theorem: equal angles subtending the same chord in the circle $OBMC$, equal angles subtending the same chord in the circle $AOMC$, and equal inscribed angles subtending arc $BA$ in the circle $AOBC$. Omitting any one of these justifications would leave a gap.
Alternative Approaches
A coordinate proof can be constructed by placing the original circle as
$$x^2+y^2=R^2,$$
taking the chord $AB$ to be a horizontal line, and assigning coordinates to $M$ on that chord. The equation of the circle through $A$, $M$, and $O$ can then be written explicitly, its second intersection with the original circle found, and the equality $MB=MC$ verified algebraically. The computations are routine but lengthy.
Another synthetic approach uses the radical axis of the two circles. The line $AC$ is the radical axis of $\omega$ and $\Gamma$, while $M$ lies on the chord $AB$. By expressing powers of $M$ with respect to the two circles and exploiting the relation obtained from $OM=OB$, one can derive $MB^2=MC^2$. The angle-chasing argument above is preferable because it avoids algebra and reveals the geometric role of the auxiliary circle through $O,B,M,C$.