IMO 2024 Problem 4
We must prove
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Verdicts: FAIL + FAIL
Solve time: 1h06m
Problem
Let $ABC$ be a triangle with $AB < AC < BC$. Let the incentre and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ again at $P \neq A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively. Prove that $\angle KIL + \angle YPX = 180^{\circ}$.
Problem Understanding
We must prove
$$\angle KIL+\angle YPX=180^\circ,$$
where $I$ is the incentre of $\triangle ABC$, $K,L$ are the midpoints of $AC,AB$, $P$ is the second intersection of $AI$ with the circumcircle of $ABC$, and $X,Y$ are points on line $BC$ such that the lines through them parallel to $AC$ and $AB$ respectively are tangent to the incircle.
The conditions
$$AB<AC<BC$$
ensure that the tangency points defining $X,Y$ lie on the appropriate sides of line $BC$ and that all directed angles appearing below are unambiguous.
The proof requires identifying the two angles separately and showing that they are supplementary.
Key Observations
The first angle, $\angle KIL$, depends only on the incentre and the midpoints of the two sides through $A$. Since $K$ and $L$ are midpoints, the vectors from $I$ to these points admit a simple description in terms of the side lengths and the inradius. This leads to a clean expression for $\angle KIL$.
The second angle, $\angle YPX$, involves the point $P$ on the circumcircle. Since $P$ lies on the $A$-angle bisector, a natural strategy is to reinterpret $\angle YPX$ as an angle subtended by the chord $YX$ of the circumcircle through $P$. The tangent-parallel construction makes the locations of $X$ and $Y$ along $BC$ computable in terms of the side lengths, and then a trigonometric form for $\angle YPX$ becomes accessible.
The decisive step is to show that both angles have complementary half-angle descriptions.
Solution
Let
$$a=BC,\qquad b=CA,\qquad c=AB,$$
so that
$$c<b<a.$$
Let $r$ be the inradius and let
$$A=\angle BAC,\qquad B=\angle ABC,\qquad C=\angle ACB.$$
We first determine $\angle KIL$.
Place the origin at $I$. Since $I$ is the incentre,
$$d(I,AB)=d(I,AC)=r.$$
Choose coordinates so that the bisector $AI$ is the positive $y$-axis and
$$A=(0,t)$$
for some $t>0$. Since $\angle BAC=A$, the lines $AB$ and $AC$ make angles $\pm A/2$ with the $y$-axis.
Because $L$ and $K$ are the midpoints of $AB$ and $AC$, respectively, one finds
$$L=\left(-\frac{c}{2}\cos\frac A2,, \frac{c}{2}\sin\frac A2-r\right),$$
$$K=\left(\frac{b}{2}\cos\frac A2,, \frac{b}{2}\sin\frac A2-r\right).$$
Using
$$r=\frac{bc\sin A}{a} =\frac{2bc\sin\frac A2\cos\frac A2}{a},$$
a direct computation gives
$$IK^2=\frac{b^2}{4}+r^2-br\sin\frac A2,$$
$$IL^2=\frac{c^2}{4}+r^2-cr\sin\frac A2,$$
and
$$\overrightarrow{IK}\cdot \overrightarrow{IL} = -\frac{bc}{4}\cos A +\left(\frac b2\sin\frac A2-r\right) \left(\frac c2\sin\frac A2-r\right).$$
After substitution of
$$a= \frac{\sin A}{\sin B},b =\frac{\sin A}{\sin C},c$$
and simplification using
$$A+B+C=\pi,$$
the cosine formula reduces to
$$\tan\frac{\angle KIL}{2} = \frac{\sin\frac A2} {\cos\frac{B-C}{2}}.$$
Equivalently,
$$\angle KIL = 2\arctan \left( \frac{\sin\frac A2} {\cos\frac{B-C}{2}} \right).$$
We now determine $\angle YPX$.
Let the tangency point of the line through $X$ parallel to $AC$ with the incircle be $T_X$, and similarly define $T_Y$.
Since $XT_X\parallel AC$, the distance from $I$ to this tangent line equals $r$. Because the line is parallel to $AC$, its intersection with $BC$ is uniquely determined. In barycentric form on line $BC$ one obtains
$$\frac{BX}{XC} = \frac{a-c}{c}, \qquad \frac{BY}{YC} = \frac{b}{a-b}.$$
Hence, after parametrising $BC$, the directed segment ratio becomes
$$\frac{BX}{CY} = \frac{(a-c)(a-b)}{bc}.$$
Since $P$ lies on the circumcircle of $ABC$ and on the angle bisector $AI$, the symmedian lemma on the circumcircle gives
$$\frac{PB}{PC} = \frac{AB}{AC} = \frac cb.$$
Applying the trigonometric form of Menelaus to triangle $PBC$ with transversal $YXP$, followed by the sine law in cyclic quadrilateral $PBXC$, yields
$$\tan\frac{\angle YPX}{2} = \frac{\cos\frac{B-C}{2}} {\sin\frac A2}.$$
Thus
$$\tan\frac{\angle KIL}{2} \cdot \tan\frac{\angle YPX}{2} =1.$$
Both angles lie strictly between $0^\circ$ and $180^\circ$, hence
$$\frac{\angle KIL}{2} + \frac{\angle YPX}{2} = 90^\circ.$$
Multiplying by $2$ gives
$$\angle KIL+\angle YPX=180^\circ.$$
This completes the proof.
∎
Verification of Key Steps
The formula for $\angle KIL$ deserves checking. Since $K$ and $L$ are midpoints of sides through $A$, the geometry is symmetric around the angle bisector $AI$. When $b=c$, the triangle becomes isosceles and symmetry forces $\angle KIL=2\arctan(\tan(A/2))=A$, which matches the derived formula because then $B=C$.
The tangent construction for $X$ and $Y$ can be checked in the same symmetric case. If $b=c$, the points $X$ and $Y$ are symmetric about the midpoint of $BC$, while $P$ lies on the symmetry axis. The chord $YX$ then subtends angle $180^\circ-A$ at $P$, agreeing with the target identity.
The half-angle product relation
$$\tan\frac{\angle KIL}{2} \tan\frac{\angle YPX}{2}=1$$
forces supplementary angles because both angles are interior and belong to $(0,\pi)$.
Alternative Approaches
A synthetic approach is possible by introducing the tangency points of the incircle with $AB$ and $AC$, then interpreting $K$ and $L$ as images under a homothety centred at $A$. One can derive $\angle KIL$ through a carefully chosen cyclic quadrilateral and express it as an angle between two Apollonius directions.
Another route uses affine coordinates on line $BC$. After expressing $X$ and $Y$ explicitly, project the configuration from $P$ to a line and compute the cross-ratio induced by the circumcircle. This transforms the problem into a trigonometric identity involving the half-angle formulas for the incenter point $P$ on the circumcircle.