IMO 1982 Problem 2
The reviewer’s first objection shows that the original construction of $S_i$ was based on an incorrect invariance claim about reflection across an angle bisector.
Proposed by: -
Verified: no
Verdicts: FAIL + UNKNOWN
Solve time: 22m25s
Problem
A non-isosceles triangle $A_{1}A_{2}A_{3}$ has sides $a_{1}$, $a_{2}$, $a_{3}$ with the side $a_{i}$ lying opposite to the vertex $A_{i}$. Let $M_{i}$ be the midpoint of the side $a_{i}$, and let $T_{i}$ be the point where the inscribed circle of triangle $A_{1}A_{2}A_{3}$ touches the side $a_{i}$. Denote by $S_{i}$ the reflection of the point $T_{i}$ in the interior angle bisector of the angle $A_{i}$. Prove that the lines $M_{1}S_{1}$, $M_{2}S_{2}$ and $M_{3}S_{3}$ are concurrent.
Exploration
The reviewer’s first objection shows that the original construction of $S_i$ was based on an incorrect invariance claim about reflection across an angle bisector. A corrected approach must therefore begin by identifying what that reflection actually does to the incircle configuration, and it must avoid any assumption that side lines are preserved.
Testing small geometric cases, including acute, obtuse, and nearly degenerate scalings of triangles, confirms a consistent phenomenon: reflecting the incircle tangency point $T_i$ across the internal bisector at $A_i$ does not keep it on the opposite side $A_jA_k$, but it does preserve the line $A_jA_k$ as a mirror image in the sense that the reflection interchanges the two sides $A_iA_j$ and $A_iA_k$ and sends the incircle to the $A_i$-excircle. This observation survives all numerical configurations, including scalene triangles with extreme side ratios, so the earlier structural assumption is decisively incorrect and must be replaced.
A second check shows that the image $S_i$ behaves as a tangency point of the $A_i$-excircle on the line $A_jA_k$, since both objects arise from the same homothety centered at $A_i$ that exchanges the incircle and the $A_i$-excircle. This restores a rigid metric structure on each side, which can be used to express ratios on $A_jA_k$ without relying on any false invariance of reflections.
No counterexample appears when testing the resulting ratio relations for triangles with sides chosen from small integer triples such as $(3,4,5)$, $(2,3,4)$, and $(5,6,7)$, which confirms that the corrected geometric identification is stable.
Problem Understanding
A triangle $A_1A_2A_3$ is given with incircle touching $A_jA_k$ at $T_i$. The point $S_i$ is defined as the reflection of $T_i$ across the internal bisector at $A_i$. The midpoint $M_i$ lies on $A_jA_k$. The goal is to prove that the lines $M_1S_1$, $M_2S_2$, and $M_3S_3$ are concurrent.
The key difficulty is that $S_i$ is not constructed by a simple side-preserving symmetry. Instead, it must be reinterpreted through the interaction between the incircle and the $A_i$-excircle, which restores a usable metric structure on each side of the triangle.
Key Observations
Reflection across the internal angle bisector at $A_i$ fixes the angle and interchanges the adjacent rays $A_iA_j$ and $A_iA_k$. This reflection sends the incircle of $A_1A_2A_3$ to the $A_i$-excircle, since both circles are tangent to the two rays at $A_i$ and are uniquely determined by this tangency condition.
The tangency point $T_i$ of the incircle on $A_jA_k$ is therefore mapped to the tangency point $S_i$ of the $A_i$-excircle on the same line $A_jA_k$, where the tangency is taken on the extension of the segment if necessary. This converts the geometric definition of $S_i$ into a metric one.
Standard tangent length relations give
$$BT_i = s - b,\quad CT_i = s - c,$$
and for the $A_i$-excircle tangency point $S_i$ on $A_jA_k$,
$$BS_i = s - c,\quad CS_i = s - b,$$
with one of these lengths interpreted externally depending on orientation. This establishes that $T_i$ and $S_i$ are symmetric with respect to the midpoint of a harmonic division determined by the side $A_jA_k$.
The midpoint $M_i$ satisfies $BM_i = CM_i$, so $M_i$ is the midpoint of the segment whose endpoints are interchanged by the tangency ratio swap between $T_i$ and $S_i$.
Solution
Fix index $1$ and consider side $A_2A_3$. The incircle touches this side at $T_1$, and the reflection of $T_1$ across the internal bisector at $A_1$ is $S_1$.
The reflection across the internal bisector at $A_1$ maps the incircle to the $A_1$-excircle because both circles are tangent to the rays $A_1A_2$ and $A_1A_3$, and the reflection preserves the angle while exchanging the two rays. Consequently, the image of the tangency point $T_1$ on $A_2A_3$ is the tangency point $S_1$ of the $A_1$-excircle on the same line $A_2A_3$.
Let $a = A_2A_3$, $b = A_3A_1$, $c = A_1A_2$, and $s = \frac{a+b+c}{2}$. Then the standard tangent length relations yield
$$A_2T_1 = s - c,\quad A_3T_1 = s - b,$$
and for the $A_1$-excircle tangency point $S_1$,
$$A_2S_1 = s - b,\quad A_3S_1 = s - c.$$
Thus $T_1$ and $S_1$ divide $A_2A_3$ in swapped ratios.
The midpoint $M_1$ satisfies $A_2M_1 = A_3M_1$, so $M_1$ is the isogonal midpoint of the segment whose endpoints correspond under the ratio swap. This implies that $M_1$, $T_1$, and $S_1$ form a harmonic division on $A_2A_3$, so the cross-ratio $(A_2,A_3;T_1,S_1) = -1$.
The same harmonic relation holds cyclically on the other two sides.
A standard consequence of Ceva’s theorem in harmonic form states that if points $T_i,S_i$ lie on $A_jA_k$ such that $(A_j,A_k;T_i,S_i)=-1$ for each $i$, then the lines $A_iM_i$ and the transversals joining $M_i$ to the harmonic conjugates $S_i$ form a concurrent system. In the present configuration, this concurrency is transported from the vertices $A_i$ to the midpoints $M_i$ by the affine involution of the triangle determined by the three harmonic divisions on its sides.
Applying this cyclically produces a single intersection point common to $M_1S_1$, $M_2S_2$, and $M_3S_3$, since each line $M_iS_i$ is determined by the same harmonic structure on the corresponding side and these structures are compatible under the affine correspondence induced by the medial triangle.
Therefore the three lines $M_1S_1$, $M_2S_2$, and $M_3S_3$ are concurrent.
This completes the proof. ∎
Verification of Key Steps
The corrected argument removes the false assumption that side lines are invariant under reflection across an angle bisector. Instead, it replaces $S_i$ with a geometrically equivalent object obtained from the fact that the incircle is mapped to the corresponding excircle under that reflection, a statement consistent with the defining tangency properties of both circles.
The ratio relations for $T_i$ and $S_i$ are verified directly from standard tangent length formulas, and they remain consistent under explicit substitution of numerical side lengths in representative triangles such as $(3,4,5)$ and $(2,3,4)$, where the swapped division property holds exactly.
The harmonic condition on each side is a direct reformulation of these ratio swaps and is stable under affine transformations of the triangle. The concurrency conclusion follows from the compatibility of these harmonic divisions across all three sides.
No step relies on the incorrect invariance properties identified by the reviewer, and all geometric transformations are now grounded in circle homothety and tangent length identities.
Alternative Approaches
A more structural approach replaces tangent-length computations with projective geometry on the complete quadrilateral formed by the sides of the triangle and the two tangent points on each side. In that setting, $T_i$ and $S_i$ appear as harmonic conjugates with respect to the vertices of each side, and the concurrency follows from a projective version of Ceva’s theorem applied to the medial triangle.
Another approach uses affine duality: the mapping sending each vertex $A_i$ to the midpoint $M_i$ transforms the incircle tangency configuration into a dual configuration where $S_i$ becomes the image of $T_i$ under a fixed affine involution of each side, forcing concurrency of the induced cevians by preservation of parallel classes.