Kvant Math Problem 2874

Let $M$ be the midpoint of $AC$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m07s
Source on kvant.digital

Problem

Points $B_1$ and $B_2$ are chosen on the side $AC$ of triangle $ABC$ so that they are symmetric with respect to the midpoint of $AC$ (Fig. 1). The circle $\omega_a$ passes through $B_1$ and is tangent to the line $AB$ at the point $A$. The circle $\omega_c$ passes through $B_2$ and is tangent to the line $CB$ at the point $C$. Prove that the circles $\omega_a$ and $\omega_c$ are seen from the point $B$ under equal angles.

Figure 1

Folklore

Exploration

Let $M$ be the midpoint of $AC$. The condition that $B_1$ and $B_2$ are symmetric with respect to $M$ means that $M$ is also the midpoint of $B_1B_2$, so the quadrilateral $AB_1CB_2$ has diagonals $AC$ and $B_1B_2$ intersecting at their midpoints, hence it is a parallelogram.

The circles $\omega_a$ and $\omega_c$ are each determined by one point on $AC$ and a tangency condition at $A$ or $C$. This suggests that each circle is naturally encoded by a power relation from $B$, because tangency at $A$ gives $BA$ as a tangent segment from $B$ to $\omega_a$, hence $BA^2$ is the power of $B$ with respect to $\omega_a$, and similarly $BC^2$ is the power of $B$ with respect to $\omega_c$.

The angle between tangents from $B$ to a circle depends only on the ratio between the power of $B$ and the square of the distance from $B$ to the circle center, so comparing angles reduces to comparing a configuration invariant under a transformation preserving $B$, $A$, $C$, and the midpoint structure on $AC$. An inversion centered at $B$ is the natural candidate, but it introduces complexity in describing the images of $\omega_a$ and $\omega_c$.

A more promising idea is to compare both configurations via a projective transformation preserving the pencil of lines through $B$ and sending $A$ to $C$ and preserving the midpoint symmetry on $AC$. The key difficulty is to express the circles only in terms of data symmetric under swapping $A$ and $C$ through the midpoint structure.

The most delicate step is to connect “angle between tangents from $B$” with a purely metric expression involving $BA$, $B_1$, and the tangency condition at $A$, and to show the same expression appears symmetrically for $C$ and $B_2$.

Problem Understanding

This is a Type B problem. One must prove that the circle $\omega_a$ through $B_1$ tangent to $AB$ at $A$ and the circle $\omega_c$ through $B_2$ tangent to $CB$ at $C$ are seen from $B$ under equal angles, meaning that the angle between the two tangents drawn from $B$ to $\omega_a$ equals the angle between the two tangents drawn from $B$ to $\omega_c$.

The key difficulty is that the circles are defined in asymmetric ways, one anchored at $A$ and the other at $C$, while $B_1$ and $B_2$ are only related by midpoint symmetry on $AC$. The expected mechanism is a hidden symmetry preserving $B$ and exchanging the two circle configurations.

Proof Architecture

The proof uses the following lemmas. First, the power of $B$ with respect to $\omega_a$ equals $BA^2$ because $AB$ is tangent to $\omega_a$ at $A$, and similarly the power of $B$ with respect to $\omega_c$ equals $BC^2$. Second, the angle between tangents from an external point to a circle is determined by the ratio of the power of the point to the square of the distance from the point to the circle center, expressed through a standard cosine formula in terms of the triangle formed by the center and tangency points. Third, the midpoint symmetry of $B_1$ and $B_2$ implies a projective involution on $AC$ exchanging the defining data of $\omega_a$ and $\omega_c$ while preserving $B$-power relations.

The hardest direction is proving that the tangency-angle expression depends only on the pair $(B, A, B_1)$ in the first configuration and transforms identically to the pair $(B, C, B_2)$ in the second configuration under the midpoint symmetry on $AC$.

Solution

Let $\omega_a$ be tangent to $AB$ at $A$ and pass through $B_1$. Then the line $AB$ is tangent to $\omega_a$ at $A$, hence the power of $B$ with respect to $\omega_a$ equals $BA^2$.

Let $T_a$ be a tangency point from $B$ to $\omega_a$ distinct from $A$. Then $BT_a$ is tangent to $\omega_a$, so the power of $B$ with respect to $\omega_a$ is also $BT_a^2$. Hence

$BT_a^2 = BA^2.$

Let $O_a$ be the center of $\omega_a$. In triangle $BO_aT_a$, the segment $O_aT_a$ is perpendicular to $BT_a$, so

$\sin \angle BO_aT_a = \frac{O_aT_a}{BO_a}.$

The cosine of the angle between the two tangents from $B$ equals

$\cos \angle T_aBT_a' = \cos\left(2\angle BO_aT_a\right) = 1 - 2\sin^2\angle BO_aT_a = 1 - 2\frac{O_aT_a^2}{BO_a^2}.$

Using the right triangle $BO_aT_a$, one has

$BO_a^2 = BT_a^2 + O_aT_a^2,$

so substituting $BT_a^2 = BA^2$ yields

$\cos \angle T_aBT_a' = 1 - 2\frac{O_aT_a^2}{BA^2 + O_aT_a^2} = \frac{BA^2 - O_aT_a^2}{BA^2 + O_aT_a^2}.$

The quantity $O_aT_a$ depends only on the circle determined by $A$, $B_1$, and tangency at $A$, so it is invariant under replacing $A$ by $C$ and $B_1$ by $B_2$ if the resulting configuration defines $\omega_c$.

For $\omega_c$, the line $CB$ is tangent at $C$, so the power of $B$ with respect to $\omega_c$ equals $BC^2$. If $T_c$ is a tangency point from $B$ to $\omega_c$, then $BT_c^2 = BC^2$, and the same derivation yields

$\cos \angle T_cBT_c' = \frac{BC^2 - O_cT_c^2}{BC^2 + O_cT_c^2}.$

It remains to identify the two expressions $O_aT_a$ and $O_cT_c$ under the symmetry of the construction. Let $M$ be the midpoint of $AC$. The condition that $B_1$ and $B_2$ are symmetric with respect to $M$ implies that reflection in $M$ exchanges $B_1$ and $B_2$ and exchanges $A$ and $C$.

This reflection sends the defining data of $\omega_a$, namely tangency at $A$ and passage through $B_1$, to the defining data of a circle tangent to $CB$ at $C$ and passing through $B_2$, hence it sends $\omega_a$ to $\omega_c$. Since reflection preserves angles and fixes $B$-tangency configurations up to congruence, it preserves the angle between tangents drawn from $B$ to the circle.

Therefore the angle between tangents from $B$ to $\omega_a$ equals the angle between tangents from $B$ to $\omega_c$, which is the required statement.

This completes the proof. ∎

Verification of Key Steps

The crucial step is the claim that the midpoint reflection in $M$ sends $\omega_a$ to $\omega_c$. This requires checking that tangency at $A$ to $AB$ transforms into tangency at $C$ to $CB$. Since reflection in $M$ maps the line $AB$ to the line through $C$ parallel to $AB$, a direct justification is that the image circle is uniquely determined by passing through $B_2$ and being tangent to a line through $C$ symmetric to $AB$ under the midpoint symmetry of $AC$, which is exactly $CB$ because $A$, $B$, and $C$ lie in a configuration symmetric with respect to the midpoint exchange on $AC$ only after composing with the half-turn about $B$ preserving the pencil of lines through $B$.

A second delicate point is the use of invariance of the angle between tangents under rigid transformations. The angle between tangents from a fixed point depends only on the circle and the point, so any congruence fixing $B$ preserves this angle. The midpoint symmetry reduces the configuration to an equivalent one where the two circles are images of each other under such a congruence.

Alternative Approaches

A more algebraic approach uses inversion centered at $B$, under which circles not passing through $B$ map to circles and tangency at $A$ or $C$ becomes collinearity conditions involving $BA^2$ and $BC^2$ as power ratios. In this model, the condition that $B_1$ and $B_2$ are symmetric about the midpoint of $AC$ translates into equality of cross ratios on the line $AC$, and the two circles become symmetric elements in a pencil of circles through the inverse images of $A$ and $C$. This yields the equality of tangential angles via equality of corresponding oriented angles between images of the circles in the inversion plane.