Kvant Math Problem 817
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Verdicts: UNKNOWN + UNKNOWN
Solve time: 7m55s
Source on kvant.digital
Problem
Point $K$ lies on the side $BC$ of triangle $ABC$. Prove that the relation $$|AK|^2=|AB|\cdot|AC|-|KB|\cdot|KC|$$ holds if and only if $|AB|=|AC|$ or $\widehat{BAK}=\widehat{CAK}$.
A. L. Toom
Exploration
Let
$$a=|AB|,\qquad b=|AC|,\qquad x=|KB|,\qquad y=|KC|,$$
so that $x+y=|BC|$.
Since $K\in BC$, Stewart's theorem immediately relates $AK$ to $a,b,x,y$:
$$a^{2}y+b^{2}x=(x+y)\bigl(|AK|^{2}+xy\bigr).$$
The given condition is
$$|AK|^{2}=ab-xy.$$
Substituting this into Stewart's formula gives
$$a^{2}y+b^{2}x=(x+y)ab.$$
After rearrangement,
$$a^{2}y-ab,y+b^{2}x-ab,x =(a-b)(ay-bx)=0.$$
This factorization is the central observation. It shows that the metric condition is equivalent to
$$(a-b)(ay-bx)=0.$$
Hence either $a=b$, or
$$\frac{x}{y}=\frac{a}{b}.$$
The second relation is exactly the conclusion of the angle bisector theorem: if $AK$ bisects $\angle BAC$, then
$$\frac{KB}{KC}=\frac{AB}{AC}.$$
Conversely, if the ratio relation holds, the converse angle bisector theorem yields
$$\angle BAK=\angle CAK.$$
The only step that could conceal an error is the passage from the original identity to
$$(a-b)(ay-bx)=0.$$
It must be checked carefully through Stewart's theorem and algebra.
Problem Understanding
We are given a point $K$ on side $BC$ of triangle $ABC$. We must prove that
$$|AK|^2=|AB|\cdot|AC|-|KB|\cdot|KC|$$
holds exactly in the two situations:
either $AB=AC$, or $AK$ is the bisector of $\angle BAC$.
This is a Type B problem, a proof of an equivalence.
The core difficulty is to convert the quadratic relation involving lengths into a geometric condition. Stewart's theorem does this conversion, reducing the statement to a factorized algebraic condition. One factor gives $AB=AC$, while the other becomes the angle-bisector condition via the angle bisector theorem.
Proof Architecture
Lemma 1. For $a=AB$, $b=AC$, $x=KB$, $y=KC$, Stewart's theorem gives
$$a^{2}y+b^{2}x=(x+y)\bigl(AK^{2}+xy\bigr).$$
This is the standard form of Stewart's theorem applied to cevian $AK$.
Lemma 2. The relation
$$AK^{2}=ab-xy$$
is equivalent to
$$(a-b)(ay-bx)=0.$$
Substitute the given relation into Lemma 1 and simplify.
Lemma 3. The equation $ay=bx$ is equivalent to
$$\angle BAK=\angle CAK.$$
Indeed,
$$\frac{x}{y}=\frac{a}{b}$$
is exactly the ratio condition of the angle bisector theorem and its converse.
The hardest direction is proving that the metric identity implies one of the two geometric alternatives. The most delicate point is the factorization in Lemma 2.
Solution
Set
$$a=|AB|,\qquad b=|AC|,\qquad x=|KB|,\qquad y=|KC|.$$
Since $K$ lies on $BC$, Stewart's theorem for cevian $AK$ gives
$$a^{2}y+b^{2}x=(x+y)\bigl(|AK|^{2}+xy\bigr). \tag{1}$$
Assume first that
$$|AK|^{2}=ab-xy. \tag{2}$$
Substituting (2) into (1), we obtain
$$a^{2}y+b^{2}x=(x+y)ab.$$
Hence
$$a^{2}y-ab,y+b^{2}x-ab,x=0.$$
Factoring,
$$(a-b)ay-(a-b)bx=0,$$
therefore
$$(a-b)(ay-bx)=0. \tag{3}$$
Equation (3) yields two possibilities.
If $a=b$, then
$$|AB|=|AC|.$$
If $ay-bx=0$, then
$$ay=bx,$$
or equivalently
$$\frac{x}{y}=\frac{a}{b}.$$
Thus
$$\frac{KB}{KC}=\frac{AB}{AC}.$$
By the converse of the angle bisector theorem,
$$\angle BAK=\angle CAK.$$
Hence the given metric relation implies that either $AB=AC$ or $AK$ bisects $\angle BAC$.
For the converse, assume that either $AB=AC$ or $\angle BAK=\angle CAK$.
If $AB=AC$, then $a=b$, and
$$(a-b)(ay-bx)=0.$$
If $\angle BAK=\angle CAK$, the angle bisector theorem gives
$$\frac{KB}{KC}=\frac{AB}{AC},$$
that is,
$$\frac{x}{y}=\frac{a}{b},$$
hence
$$ay-bx=0.$$
Again,
$$(a-b)(ay-bx)=0.$$
Expanding this equality,
$$a^{2}y+b^{2}x=(x+y)ab. \tag{4}$$
Comparing (4) with Stewart's identity (1),
$$(x+y)\bigl(|AK|^{2}+xy\bigr)=(x+y)ab.$$
Since $x+y=BC>0$, division by $x+y$ yields
$$|AK|^{2}+xy=ab.$$
Therefore
$$|AK|^{2}=ab-xy,$$
or, in the original notation,
$$|AK|^{2}=|AB|\cdot|AC|-|KB|\cdot|KC|.$$
The implication in both directions has been proved. This completes the proof.
∎
Verification of Key Steps
The first delicate step is the algebra after substitution into Stewart's theorem. Starting from
$$a^{2}y+b^{2}x=(x+y)ab,$$
we move all terms to one side:
$$a^{2}y-ab,y+b^{2}x-ab,x.$$
Grouping terms gives
$$ay(a-b)+bx(b-a),$$
which equals
$$(a-b)(ay-bx).$$
No sign changes are omitted.
The second delicate step is the conversion of $ay=bx$ into an angle statement. From
$$ay=bx$$
we obtain
$$\frac{KB}{KC}=\frac{AB}{AC}.$$
Because $K$ lies on side $BC$, the converse angle bisector theorem applies directly and yields
$$\angle BAK=\angle CAK.$$
Without the condition $K\in BC$, this conclusion would not be justified.
A separate verification comes from testing the two special cases. If $a=b$, then Stewart's theorem becomes
$$a^{2}(x+y)=(x+y)(AK^{2}+xy),$$
so
$$AK^{2}=a^{2}-xy=ab-xy.$$
If $AK$ is an angle bisector, then $x:y=a:b$; writing $x=at$, $y=bt$ and substituting into Stewart's theorem gives
$$ab(x+y)=(x+y)(AK^{2}+xy),$$
hence again
$$AK^{2}=ab-xy.$$
Alternative Approaches
A different proof can be obtained from the law of cosines in triangles $ABK$ and $ACK$. Writing
$$AB^{2}=AK^{2}+KB^{2}-2,AK\cdot KB\cos\angle BKA,$$
and
$$AC^{2}=AK^{2}+KC^{2}-2,AK\cdot KC\cos\angle AKC,$$
then using $\angle BKA+\angle AKC=180^\circ$, one may eliminate the cosine terms and derive an equation equivalent to
$$(a-b)(ay-bx)=0.$$
The remainder proceeds through the angle bisector theorem.
The Stewart-theorem approach is preferable because the given condition already involves the quantities appearing naturally in Stewart's formula. The factorization emerges immediately and leads directly to the geometric interpretation.