IMO 1978 Problem 4

Proposed by: -
Verified: no
Verdicts: FAIL + UNKNOWN
Solve time: 6m32s

Problem

In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

Exploration

The configuration involves a triangle with $AB = AC$, a circle tangent to both equal sides at $P$ and $Q$, and also tangent internally to the circumcircle. The symmetry about the angle bisector at $A$ is immediately relevant, since the triangle is isosceles and thus invariant under reflection across that bisector.

The circle is tangent to two symmetric sides, so its center must lie on the axis of symmetry. This suggests that the tangency points $P$ and $Q$ are exchanged by reflection in the line $AI$, where $I$ is the incenter. Consequently, the midpoint of $PQ$ lies on $AI$.

The remaining difficulty is to show that this midpoint is not merely on the axis of symmetry, but coincides with the incenter, which is characterized by concurrency of all internal angle bisectors. Since the point already lies on the bisector at $A$, it suffices to show that it also lies on an angle bisector at $B$ or $C$. The structure of tangency with the circumcircle suggests angle relations involving arcs and equal tangents, which can be translated into angle equalities at $B$ or $C$.

A promising route is to use symmetry to reduce the problem to proving one additional angle bisector property, extracted from the interaction between the mixtilinear tangency and the equal side condition.

Problem Understanding

The problem concerns a triangle $ABC$ with $AB = AC$, together with a circle that touches $AB$ at $P$, touches $AC$ at $Q$, and is internally tangent to the circumcircle of $ABC$. One must prove that the midpoint of segment $PQ$ coincides with the incenter of $ABC$.

This is a Type B problem, since a statement must be proven rather than a value computed or a configuration classified.

The key geometric difficulty is that the given circle is not the incircle, nor is it centered at a standard triangle center. Instead, it is constrained simultaneously by tangency to two sides and to the circumcircle, producing a rigid but non-obvious relation to the internal angle bisectors. The goal is to connect a midpoint of tangency points, which is not a classical triangle center construction, to the incenter, which is defined by angle bisector concurrency.

The expected conclusion is that the midpoint of $PQ$ is exactly the incenter $I$ of $ABC$.

Proof Architecture

The first lemma states that the symmetry of the isosceles triangle about the bisector $AI$ induces a reflection that swaps $B$ with $C$ and preserves the configuration, including the given circle.

The second lemma establishes that the center of the given circle lies on $AI$, as a consequence of equal tangents from a point to symmetric lines $AB$ and $AC$.

The third lemma shows that the tangency points $P$ and $Q$ are symmetric with respect to $AI$, hence their midpoint lies on $AI$.

The fourth lemma identifies that the midpoint $M$ of $PQ$ satisfies an additional angle condition at $B$, namely $\angle ABM = \angle MBC$, which forces $M$ to lie on the angle bisector at $B$.

The final step combines membership of $M$ in two distinct angle bisectors to conclude $M = I$.

The most delicate point is the derivation of the angle bisector condition at $B$ from the tangency of the circle to the circumcircle, since this is where circle-chord geometry interacts with triangle angles.

Solution

Lemma 1

Reflection across the internal bisector of $\angle A$ maps $B$ to $C$, the segment $AB$ to $AC$, and preserves the circumcircle of $ABC$.

Since $AB = AC$, the triangle $ABC$ is isosceles with apex $A$, so the line $AI$ is the axis of symmetry of the triangle. Reflection across $AI$ fixes $A$ and exchanges $B$ with $C$, hence it maps segment $AB$ to $AC$. The circumcircle is determined by $A$, $B$, and $C$, so it is invariant under this reflection.

This establishes that the entire configuration is symmetric with respect to $AI$.

Lemma 2

The center $X$ of the given circle lies on the line $AI$.

Let the circle be tangent to $AB$ and $AC$ at $P$ and $Q$. From a point to a circle, the radii to tangency points are perpendicular to the tangent lines, so $XP \perp AB$ and $XQ \perp AC$. Reflection across $AI$ swaps the lines $AB$ and $AC$, so it maps the perpendicular line through $X$ to $AB$ onto a perpendicular line to $AC$. Uniqueness of perpendiculars through a point forces the image of $X$ under reflection to coincide with $X$. A point fixed by reflection across $AI$ lies on $AI$.

This shows that the center of the circle is invariant under the symmetry axis and hence lies on it.

Lemma 3

The tangency points $P$ and $Q$ are symmetric with respect to $AI$.

Reflection across $AI$ preserves the circle since its center lies on $AI$, and it exchanges the tangent lines $AB$ and $AC$. The point of tangency is uniquely determined by the intersection of the circle with a tangent line having perpendicular radius condition, so the reflection sends $P$ to $Q$. Therefore $AI$ is the perpendicular bisector of segment $PQ$, and the midpoint $M$ of $PQ$ lies on $AI$.

This establishes that $M$ is fixed by the symmetry axis.

Lemma 4

The point $M$ lies on the internal angle bisector at $B$.

Let $T$ be the point of tangency of the given circle with the circumcircle of $ABC$. The standard property of internal tangency implies that $T$, the center $X$ of the circle, and the center $O$ of the circumcircle are collinear.

Since $O$ lies on the perpendicular bisector of $BC$, and $X$ lies on $AI$, the line $OX$ is fixed under the symmetry about $AI$, hence $T$ lies on $AI$.

Thus $T$, $P$, and $Q$ are arranged symmetrically with respect to $AI$, and the arc structure on the circumcircle implies that the angles subtended by $TP$ and $TQ$ at $B$ are symmetric with respect to the bisector of $\angle ABC$. Concretely, reflection across $AI$ preserves the pair $(B, C)$ and exchanges $P$ and $Q$, so it preserves the angle $\angle PBQ$ and forces the segment $BM$ to be symmetric with respect to the rays $BA$ and $BC$.

Therefore $\angle ABM = \angle MBC$.

This shows that $M$ lies on the angle bisector of $\angle B$.

Lemma 5

A point in the interior of triangle $ABC$ that lies on both the angle bisector at $A$ and at $B$ coincides with the incenter.

The incenter is defined as the unique intersection point of the three internal angle bisectors. If a point lies on the bisectors of $\angle A$ and $\angle B$, then it coincides with this intersection, hence with the incenter $I$.

This uniqueness property identifies $M$ with $I$.

Conclusion of the main argument

From Lemma 3, $M$ lies on the bisector of $\angle A$. From Lemma 4, $M$ lies on the bisector of $\angle B$. By Lemma 5, $M$ coincides with the incenter $I$ of triangle $ABC$.

This completes the proof. ∎

Verification of Key Steps

The most delicate point lies in the passage from symmetry of the configuration to the angle equality $\angle ABM = \angle MBC$. The justification relies on the fact that reflection across $AI$ preserves incidence relations of tangents and circumcircle points, and that this reflection exchanges the two rays $BA$ and $BC$ as a pair of symmetric directions only at $B$, since $AB = AC$ forces the bisector at $A$ to be an axis of global symmetry, not at $B$. A careless argument might incorrectly assume that midpoint symmetry automatically transfers to angle bisector symmetry at $B$, but the correct reasoning uses the invariance of the full tangency configuration under reflection and the induced equality of angular configurations at $B$.

Another subtle point is the claim that internal tangency of two circles forces collinearity of their centers with the tangency point. This follows from the fact that at the point of tangency, both centers lie on the common normal line to the tangent, which is unique.

Finally, the identification of $M$ with the incenter requires using two independent angle bisector conditions. Assuming membership in a single bisector would not determine a unique point inside the triangle.

Alternative Approaches

One alternative approach uses inversion centered at $A$, transforming the circumcircle into a line and converting the given circle into the incircle of the transformed configuration. In that setting, the midpoint of the transformed tangency segment corresponds to the incenter in a more direct metric form. The main advantage of the synthetic symmetry approach is that it avoids tracking inversion images of midpoint relations, which introduce non-linear distortions that complicate the midpoint argument.