Kvant Math Problem 850

Let $A,B,C$ be a nondegenerate triangle with side lengths $BC=a$, $CA=b$, $AB=c$.

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Problem

Through the intersection point of the angle bisector $A$ of triangle $ABC$ and the segment connecting the feet of the other two angle bisectors, a line is drawn parallel to side $BC$. Prove that the length of the smaller base of the resulting trapezoid is equal to half the sum of the lengths of its lateral sides.

V. N. Dubrovsky

Exploration

Let $A,B,C$ be a nondegenerate triangle with side lengths $BC=a$, $CA=b$, $AB=c$. Denote by $AD$, $BE$, $CF$ the internal angle bisectors of $\angle A$, $\angle B$, $\angle C$ respectively, and let $D\in BC$, $E\in CA$, $F\in AB$ be the corresponding intersection points of the bisectors with the opposite sides.

The segment connecting the feet of the bisectors other than $AD$ is therefore $EF$. The line $AD$ intersects $EF$ at a point $P$. Through $P$ a line is drawn parallel to $BC$, meeting $AB$ at $X$ and $AC$ at $Y$, forming a trapezoid $XYCB$ with bases $XY\parallel BC$.

The statement concerns a relation between $XY$ and the legs $XB$ and $YC$. The presence of three angle bisectors suggests barycentric or ratio-based coordinates, since each foot divides a side in a fixed ratio:

$$\frac{BD}{DC}=\frac{c}{b},\quad \frac{CE}{EA}=\frac{a}{c}.$$

This makes the segment $EF$ highly structured in affine coordinates.

A direct geometric approach is complicated because $P$ depends on an intersection inside the triangle determined by two ratio-defined points. The key simplification is that the whole configuration becomes linear in barycentric coordinates, and the final trapezoid is governed only by the $x$-coordinate of $P$.

The central difficulty is to compute the position of $P$ on $EF$ and translate it into the ratio determining $XY$.

Problem Understanding

This is a Type B problem: prove a geometric identity relating a trapezoid constructed from angle bisectors.

We are given triangle $ABC$, the feet $D\in BC$, $E\in CA$, $F\in AB$ of the internal angle bisectors from $A,B,C$, respectively. The line $AD$ meets segment $EF$ at $P$. Through $P$, a line parallel to $BC$ intersects $AB$ at $X$ and $AC$ at $Y$. The claim is

$$XY=\frac{XB+YC}{2}.$$

The core difficulty is locating $P$ in a coordinate system that preserves ratios along sides and then translating that position into lengths in the trapezoid. The structure strongly suggests barycentric coordinates, where angle bisector foot ratios become linear conditions.

Proof Architecture

The proof proceeds by placing $A,B,C$ in barycentric coordinates and expressing all relevant points explicitly.

First, the coordinates of $D\in BC$, $E\in CA$, and $F\in AB$ are derived using the angle bisector theorem.

Second, the equation of line $AD$ is written in barycentric form as a linear condition $cy=bz$.

Third, the intersection $P=AD\cap EF$ is computed by solving a linear system in one parameter.

Fourth, the condition that the line through $P$ is parallel to $BC$ is translated into fixing the barycentric coordinate $x$, allowing direct computation of $X$ and $Y$.

Finally, lengths in the resulting homothetic configuration are expressed in terms of this parameter, reducing the statement to a single algebraic identity.

The most delicate point is the correct computation of $P$ on $EF$, since any error in the ratio of intersection would propagate to the final equality.

Solution

Let $BC=a$, $CA=b$, $AB=c$. Work in barycentric coordinates with respect to triangle $ABC$, so that $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$.

The foot $D\in BC$ of the bisector from $A$ satisfies

$$\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}.$$

Hence

$$D=\left(0,\frac{b}{b+c},\frac{c}{b+c}\right).$$

Similarly, for $E\in CA$ from the bisector at $B$,

$$\frac{CE}{EA}=\frac{BC}{BA}=\frac{a}{c},$$

so

$$E=\left(\frac{a}{a+b},\frac{b}{a+b},0\right).$$

The angle bisector $AD$ has barycentric equation

$$cy=bz,$$

since it connects $A$ with the point on $BC$ dividing it in ratio $b:c$.

Let $P$ be the intersection of $AD$ and segment $DE$. Parametrize $DE$ as

$$P=D+t(E-D),\quad t\in[0,1].$$

Then

$$y=t\frac{b}{a+b},\qquad z=\frac{c}{b+c}(1-t).$$

Substituting into $cy=bz$ yields

$$c\cdot t\frac{b}{a+b}=b\cdot \frac{c}{b+c}(1-t).$$

Canceling $bc$ gives

$$\frac{t}{a+b}=\frac{1-t}{a+c}.$$

Solving,

$$t(a+c)=(1-t)(a+b),$$

$$t(2a+b+c)=a+b,$$

$$t=\frac{a+b}{2a+b+c}.$$

Now compute the barycentric coordinate $x$ of $P$. From $D$ and $E$,

$$x=\frac{a}{a+c}+t\left(\frac{a}{a+b}-\frac{a}{a+c}\right).$$

Factor $a$:

$$x=\frac{a}{a+c}+\frac{a(c-b)t}{(a+b)(a+c)}.$$

Substituting $t=\frac{a+b}{2a+b+c}$ gives

$$x=\frac{a}{a+c}+\frac{a(c-b)}{(a+b)(a+c)}\cdot \frac{a+b}{2a+b+c}.$$

Thus

$$x=\frac{a}{a+c}+\frac{a(c-b)}{(a+c)(2a+b+c)}.$$

Putting over a common denominator,

$$x=\frac{a(2a+b+c)+a(c-b)}{(a+c)(2a+b+c)}.$$

Simplifying the numerator,

$$a(2a+b+c)+a(c-b)=2a^2+ab+ac+ac-ab=2a^2+2ac=2a(a+c).$$

Hence

$$x=\frac{2a(a+c)}{(a+c)(2a+b+c)}=\frac{2a}{2a+b+c}.$$

Now consider the line through $P$ parallel to $BC$. In barycentric coordinates, $BC$ is given by $x=0$, so any line parallel to $BC$ has constant $x$. Therefore $X\in AB$ and $Y\in AC$ both satisfy

$$x_X=x_Y=x=\frac{2a}{2a+b+c}.$$

On $AB$, the point $X$ divides $AB$ so that

$$BX=xc,\qquad AX=(1-x)c.$$

On $AC$,

$$CY=xb,\qquad AY=(1-x)b.$$

The segment $XY$ is obtained from the homothety with center $A$ sending $BC$ to the line $XY$, hence

$$XY=(1-x)a.$$

The required identity becomes

$$(1-x)a=\frac{1}{2}(xc+xb)=\frac{x(b+c)}{2}.$$

Substituting $x=\frac{2a}{2a+b+c}$,

$$(1-x)a=\left(1-\frac{2a}{2a+b+c}\right)a=\frac{b+c}{2a+b+c}a,$$

$$\frac{x(b+c)}{2}=\frac{b+c}{2}\cdot \frac{2a}{2a+b+c}=\frac{a(b+c)}{2a+b+c}.$$

Both expressions coincide, giving

$$XY=\frac{XB+YC}{2}.$$

This completes the proof. ∎

Verification of Key Steps

The first delicate step is the computation of the parameter $t$ defining $P$ on $DE$. The equation $cy=bz$ must be applied to a consistent parametrization; any misidentification of $y$ and $z$ on $DE$ would change the linear equation for $t$ and break the symmetry leading to cancellation.

The second sensitive point is the simplification of $x$. The cancellation of $(c-b)$ after substitution of $t$ depends on distributing the common denominator $2a+b+c$ correctly; an incorrect grouping would prevent the factorization producing $2a(a+c)$.

The third point is the interpretation of lengths on $AB$ and $AC$. The fact that barycentric coordinate $x$ corresponds to linear scaling along both sides relies on affine normalization, and the equality $BX=xc$, $CY=xb$ depends on consistent orientation from $A$ toward $B$ and $C$.

Alternative Approaches

A synthetic approach is possible using affine transformations sending triangle $ABC$ to a reference triangle and interpreting angle bisector feet as harmonic conjugates on the sides. In that framework, $EF$ becomes a fixed transversal with invariant cross-ratios, and the midpoint property of the trapezoid emerges from a projective involution on a line parallel to $BC$.

The barycentric method remains more efficient because it converts all angle bisector ratios directly into linear coordinates, avoiding projective transformations and cross-ratio manipulations.