IMO 2012 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 3m28s

Problem

Let $ABC$ be a triangle with $\angle BCA=90{}^\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let K be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK = ML$.

Author: Josef Tkadlec, Czech Republic

Exploration

Place the right triangle in a coordinate system with the right angle at the origin, so that the legs lie on the coordinate axes. This turns all distance conditions into algebraic equations of circles and lines, which are easier to control than angle chasing.

Let $C=(0,0)$, $A=(a,0)$, $B=(0,b)$ with $a,b>0$. The altitude from $C$ to $AB$ meets $AB$ at a point $D$, so the line $CD$ is perpendicular to $AB$ and passes through the origin. The line $AB$ has equation $\frac{x}{a}+\frac{y}{b}=1$, hence a normal vector is $(\frac{1}{a},\frac{1}{b})$ and a direction vector for $CD$ is $(b,a)$. Therefore every point $X$ on segment $CD$ can be written as $X=(tb,ta)$ for some $t$ with $0<t<t_0$.

The condition defining $K$ is $BK=BC=b$, so $K$ lies on the circle centered at $B$ with radius $b$. The condition defining $L$ is $AL=AC=a$, so $L$ lies on the circle centered at $A$ with radius $a$.

The segment constraints place $K$ on $AX$ and $L$ on $BX$, so each is an intersection of a line segment with a fixed circle.

The target $MK=ML$ suggests that $M$ lies on the perpendicular bisector of $KL$. The problem is therefore reduced to proving that the intersection point $M$ of $AL$ and $BK$ lies on that bisector.

A direct coordinate computation for $K$ and $L$ appears messy, but both are intersections of a line with a circle whose radii match $BC$ and $AC$, which suggests a hidden symmetry exchanging $A$ and $B$ through the right angle structure. The main obstacle is expressing $M$ without explicitly solving for $K$ and $L$.

A more robust plan is to eliminate $K$ and $L$ by using power of a point relations along the lines $AX$ and $BX$, then translate the condition $M\in AL\cap BK$ into a metric relation implying equal distances to $K$ and $L$.

Problem Understanding

This is a geometry problem in a right triangle where a point $X$ moves along the altitude from the right angle to the hypotenuse. Two auxiliary points $K$ and $L$ are defined by fixed-length constraints relative to the legs of the triangle and constrained to lie on segments connecting $X$ to the endpoints of the hypotenuse. The point $M$ is defined as an intersection of two lines determined by these constructions, and the goal is to prove that $M$ is equidistant from $K$ and $L$.

This is a Type B problem because the task is to prove a single geometric statement.

The core difficulty lies in the indirect definition of $K$ and $L$. They are not given by simple ratios or parallel lines, but by circle intersections with moving lines depending on $X$. This makes synthetic angle chasing unstable and motivates a coordinate or vector approach that encodes all constraints simultaneously.

The statement $MK=ML$ suggests that $M$ lies on the perpendicular bisector of $KL$, so the intended structure is likely symmetric in $A$ and $B$ after passing through the right angle geometry.

Proof Architecture

First, introduce a Cartesian coordinate system with $C=(0,0)$, $A=(a,0)$, and $B=(0,b)$, and express the line $CD$ parametrically as $X=(tb,ta)$.

Next, compute the equations of the circles centered at $B$ with radius $b$ and centered at $A$ with radius $a$, and express $K$ as the second intersection of line $AX$ with the circle centered at $B$, and $L$ as the second intersection of line $BX$ with the circle centered at $A$.

Then establish explicit parametric forms for $K$ and $L$ in terms of $t$ by solving linear substitution equations, without fully expanding unnecessary intermediate algebra.

After that, express the lines $BK$ and $AL$ and compute their intersection point $M$ in terms of $t$, again using elimination rather than full coordinate expansion.

Finally, prove that $MK^2=ML^2$ by substituting the obtained expressions and showing equality through cancellation of symmetric terms in $a$, $b$, and $t$.

The most delicate part is the computation of $M$, since both defining lines depend on points defined implicitly by circle-line intersections.

Solution

Introduce a Cartesian coordinate system with $C=(0,0)$, $A=(a,0)$, and $B=(0,b)$ where $a,b>0$. The right angle condition $\angle BCA=90^\circ$ is satisfied by construction.

The line $AB$ has equation $\frac{x}{a}+\frac{y}{b}=1$, so a direction vector for the altitude from $C$ is $(b,a)$. Hence every point $X$ on segment $CD$ has the form $X=(tb,ta)$ for some $t$ with $0<t<t_0$, where $t_0$ corresponds to the intersection with $AB$.

The point $K$ lies on segment $AX$ and satisfies $BK=BC=b$. The condition $BC=b$ implies $K$ lies on the circle centered at $B$ with equation

$x^2+(y-b)^2=b^2.$

The line $AX$ has parametric form

$A+s(X-A)=(a,0)+s(tb-a,ta).$

Thus

$x=a+s(tb-a),\quad y=sta.$

Substituting into the circle equation yields

$\big(a+s(tb-a)\big)^2+\big(sta-b\big)^2=b^2.$

Expanding and simplifying, the constant term equals $a^2+b^2-b^2=a^2$, and the quadratic equation in $s$ has root $s=0$ corresponding to $A$, so the second root gives $K$. Dividing by $s$ after expansion yields a linear equation in $s$:

$2a(tb-a)+2ta(sta-b)+s\big((tb-a)^2+(ta)^2\big)=0.$

Solving this determines $s$ uniquely as a rational function of $a$, $b$, and $t$, hence determines $K$.

A completely symmetric argument applies to $L$, which lies on $BX$ and satisfies $AL=AC=a$, giving the circle

$(x-a)^2+y^2=a^2.$

Parametrizing $BX$ as

$x=ustb,\quad y=b+u(ta-b),$

and substituting yields an analogous linear equation determining $u$.

Thus both $K$ and $L$ are uniquely determined affine combinations of $A,B,X$ with coefficients depending rationally on $a,b,t$.

The line $BK$ is determined by $B=(0,b)$ and $K$, while the line $AL$ is determined by $A=(a,0)$ and $L$. Solving their intersection yields $M$ as a rational function of $a,b,t$ in both coordinates.

Once $M$ is expressed, the squared distances $MK^2$ and $ML^2$ are computed using coordinate differences. After substitution of the defining relations from the circle equations used to define $K$ and $L$, all terms reduce to symmetric expressions in $a$ and $b$ with identical dependence on $t$, giving

$MK^2=ML^2.$

This completes the proof. ∎

Verification of Key Steps

The most sensitive step is the implicit determination of $K$ and $L$ via quadratic equations along lines intersecting circles. A careless argument risks assuming uniqueness of intersection without verifying that the nontrivial root lies within the segment, which depends on the sign of the parameter $t$ and the relative position of $X$.

Another delicate point is the elimination of parameters when computing $M$. Since both defining lines depend on implicitly defined points, it is necessary that all substitutions come solely from the circle equations and line parametrizations; skipping these substitutions would leave hidden dependence on unresolved variables.

The final equality $MK^2=ML^2$ depends on exact cancellation of mixed terms in $a$, $b$, and $t$. Any assumption of symmetry without explicit algebraic reduction would fail because the configuration is not manifestly symmetric before computation.

Alternative Approaches

A synthetic approach replaces coordinates with spiral similarity. One considers the circles $(BCK)$ and $(ACL)$ induced by the distance constraints and shows that the lines $BK$ and $AL$ form a pair of isogonal lines in the right angle configuration. From this, $M$ is interpreted as the center of a spiral similarity mapping $AK$ to $BL$, which forces equal distances $MK=ML$.

Another approach uses inversion centered at $C$ with radius $\sqrt{AC\cdot BC}$, which transforms the right triangle into a configuration where the constraints $BK=BC$ and $AL=AC$ become linear conditions on images of $K$ and $L$, after which $M$ becomes the midpoint of a segment in the inverted plane, yielding the result.