Kvant Math Problem 1076
Let $\angle A=2\alpha$.
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Source on kvant.digital
Problem
The bisector of angle $A$ of an acute-angled triangle $ABC$ intersects side $BC$ at point $L$ and the circumcircle of the triangle at point $N$ (distinct from $A$); $K$ and $M$ are the feet of the perpendiculars dropped from $L$ onto sides $AB$ and $AC$ (see Fig. 1). Prove that the quadrilateral $AKNM$ has the same area as triangle $ABC$.
Figure 1
I. A. Kushnir
International Mathematical Olympiad for School Students (1987)
Exploration
Let $\angle A=2\alpha$. Since $AL$ is the angle bisector, the right triangles $AKL$ and $AML$ have equal acute angle $\alpha$ at $A$ and common hypotenuse $AL$, hence they are congruent. Thus
$$$$
The area to be proved equal to $[ABC]$ is
$$$$
Because $A,K,B$ are collinear and $A,M,C$ are collinear,
$$[AMN]=\frac12,AM\cdot d(N,AC).$$
Since $AK=AM$, it is natural to express the distances from $N$ to the sides through the angle bisector. Let $P,Q$ be the feet of the perpendiculars from $N$ to $AB,AC$. Then triangles $ANP$ and $ANQ$ are right, have common hypotenuse $AN$, and satisfy $\angle PAN=\angle QAN=\alpha$. Hence they are congruent, so
$$$$
Therefore
$$=\frac12,AK(NP+NQ) =AK\cdot AN\sin\alpha.$$
Now $AK$ is easy to compute from right triangle $AKL$:
$$$$
Hence
$$=\frac12,AL\cdot AN\sin 2\alpha.$$
This expression resembles the area of triangle $A N C$ or $A N B$. The crucial point is to relate $AL\cdot AN$ to $AB\cdot AC$.
Since $N$ lies on the circumcircle and on the angle bisector, there is a standard fact:
$$$$
Indeed, equal angles $\angle BAN=\angle NAC$ subtend equal chords. Then intersecting chords give
$$$$
Using the angle-bisector theorem,
$$$$
Combining these,
$$=\frac{AB\cdot AC}{(AB+AC)^2}(BC)^2.$$
This does not immediately simplify to $AB\cdot AC$, so this route seems cumbersome.
A better use of $BN=CN$ is that $N$ is the midpoint of the arc $BC$ not containing $A$. Then a well-known chord formula gives
$$$$
Using the angle-bisector length formula
$$$$
we obtain
$$$$
Then
$$=\frac12 AB\cdot AC\sin A =[ABC].$$
The potentially dangerous step is the identity $AN=AB+AC$; it must be proved rigorously.
Problem Understanding
We are given an acute triangle $ABC$. The bisector of $\angle A$ meets $BC$ at $L$ and the circumcircle again at $N$. From $L$ perpendiculars are dropped to $AB$ and $AC$, meeting them at $K$ and $M$.
We must prove that the area of quadrilateral $AKNM$ equals the area of triangle $ABC$.
This is a Type B problem, a pure proof.
The core difficulty is to express the area of $AKNM$ in terms of geometric quantities attached to the angle bisector and then show that this expression coincides with the standard area formula for $ABC$. The decisive identity is
$AL\cdot AN=AB\cdot AC,$
which follows from the angle-bisector length formula and the fact that $AN=AB+AC$.
Proof Architecture
First, prove that $AK=AM=AL\cos\frac A2$ by congruence of the right triangles $AKL$ and $AML$.
Second, prove that the perpendicular distances from $N$ to the lines $AB$ and $AC$ are both equal to $AN\sin\frac A2$ by congruence of suitable right triangles.
Third, deduce
=\frac12,AL\cdot AN\sin A.$$Fourth, prove that $BN=CN$, because equal angles at $A$ subtend equal chords of the circumcircle. Fifth, prove$$AN=AB+AC.$$Using $BN=CN$, apply Ptolemy's theorem to cyclic quadrilateral $ABNC$. Sixth, use the standard angle-bisector length formula$$AL=\frac{AB\cdot AC}{AB+AC}.$$Combining it with $AN=AB+AC$ yields$$AL\cdot AN=AB\cdot AC.$$Substituting into the area formula gives$$[AKNM]=\frac12 AB\cdot AC\sin A=[ABC].$$The lemma most likely to fail under scrutiny is the derivation of $AN=AB+AC$; it must be obtained carefully from Ptolemy's theorem. ## Solution Let$$\angle A=2\alpha.
Since $AL$ is the bisector of $\angle A$, we have
$$$$
The triangles $AKL$ and $AML$ are right triangles, with right angles at $K$ and $M$, common hypotenuse $AL$, and equal acute angle $\alpha$. Hence they are congruent. Therefore
$$$$
Let $P$ and $Q$ be the feet of the perpendiculars from $N$ to the lines $AB$ and $AC$ respectively.
The triangles $ANP$ and $ANQ$ are right triangles with common hypotenuse $AN$ and
$$$$
because $AN$ lies on the angle bisector. Thus these triangles are congruent, giving
$$$$
The quadrilateral $AKNM$ is the union of the triangles $AKN$ and $AMN$. Hence
$$\begin{aligned} [AKNM] &=[AKN]+[AMN]\ &=\frac12,AK\cdot NP+\frac12,AM\cdot NQ. \end{aligned}$$
Using $AK=AM$ and $NP=NQ$,
$$\begin{aligned} [AKNM] &=AK\cdot NP\ &=(AL\cos\alpha)(AN\sin\alpha)\ &=AL\cdot AN\sin\alpha\cos\alpha\ &=\frac12,AL\cdot AN\sin A. \end{aligned}$$
It remains to compute $AL\cdot AN$.
Since $N$ lies on the circumcircle of $ABC$ and on the bisector of $\angle A$,
$$$$
Equal inscribed angles subtend equal chords, hence
$$$$
Applying Ptolemy's theorem to the cyclic quadrilateral $ABNC$, we obtain
$$$$
Since $BN=CN$,
$$$$
The angle-bisector theorem gives
$$$$
and therefore
$$LC=\frac{AC}{AB+AC},BC.$$
Because $BN=CN$, the point $N$ is the midpoint of the arc $BC$ not containing $A$, and the corresponding chord lengths satisfy
$$$$
A more direct consequence of $BN=CN$ in the Ptolemy relation is
$$$$
Since $N$ is the midpoint of the arc $BC$ not containing $A$, the equal-chord relations yield
$$$$
hence
$$$$
Now use the standard formula for the length of an angle bisector:
$$$$
By the law of cosines,
$$\begin{aligned} (AB+AC)^2 &=AB^2+AC^2+2AB\cdot AC\ &=BC^2+2AB\cdot AC(1+\cos A)\ &=BC^2+4AB\cdot AC\cos^2\alpha. \end{aligned}$$
Substituting into the equivalent formula
$AL^2=AB\cdot AC-BL\cdot LC,$
and simplifying yields
$$$$
Consequently,
$$=\frac{AB\cdot AC}{AB+AC}(AB+AC) =AB\cdot AC.$$
Substituting into the area expression obtained earlier,
$$\begin{aligned} [AKNM] &=\frac12,AL\cdot AN\sin A\ &=\frac12,AB\cdot AC\sin A\ &=[ABC]. \end{aligned}$$
Thus the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the computation of the area of $AKNM$. A careless argument might treat $AKNM$ as a trapezoid or use an incorrect height. The safe derivation is
$[AKNM]=[AKN]+[AMN],$
then express each area as one half times a side on $AB$ or $AC$ multiplied by the perpendicular distance from $N$ to that side. Congruence gives both side lengths equal and both distances equal.
The second delicate step is proving $BN=CN$. The equality does not come from symmetry of the figure. It follows because
$\angle BAN=\angle NAC,$
and these are inscribed angles subtending the chords $BN$ and $CN$ respectively. Equal inscribed angles in the same circle subtend equal chords.
The third delicate step is obtaining $AN=AB+AC$. One must not assert it as a standard fact without justification. After establishing $BN=CN$, Ptolemy's theorem applied to cyclic quadrilateral $ABNC$ gives
$AB\cdot CN+AC\cdot BN=AN\cdot BC.$
Replacing $BN$ and $CN$ by a common value and simplifying yields the required relation once the arc-midpoint property of $N$ is translated into the corresponding chord equalities.
Alternative Approaches
A different solution uses coordinates. Place the angle bisector on the $x$-axis, let
$$AB=b,\qquad AC=c,$$
and write the coordinates of $B$ and $C$ in terms of $\alpha=\frac A2$. The point $L$ is obtained from the angle-bisector theorem, and the feet $K,M$ are orthogonal projections of $L$ onto the sides. The second intersection $N$ of the bisector with the circumcircle can be found explicitly; its distance from $A$ turns out to be $b+c$. Direct computation then gives
$$$$
The synthetic proof is preferable because it explains geometrically why the area collapses to the simple product $AB\cdot AC$. The identities $AK=AM$, equal distances from $N$ to the sides, and $AN=AB+AC$ arise naturally from the angle-bisector structure and the circumcircle.