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1987 IMO Problems/Problem 2: Difference between revisions

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<math>\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA</math>
<math>\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA</math>


<math>\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C\frac{1}{2}A)+BM\sin (C\frac{1}{2}A)</math>
<math>\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)</math>


<math>\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)</math>
<math>\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)</math>

Revision as of 22:05, 24 January 2009

Problem

In an acute-angled triangle $ABC$ the interior bisector of the angle $A$ intersects $BC$ at $L$ and intersects the circumcircle of $ABC$ again at $N$. From point $L$ perpendiculars are drawn to $AB$ and $AC$, the feet of these perpendiculars being $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.

Solution

We are to prove that $[AKNM]=[ABC]$ or equivilently, $[ABC]+[BNC]-[KNC]-[BMN]=[ABC]$. Thus, we are to prove that $[BNC]=[KNC]+[BMN]$. It is clear that since $\angle BAN=\angle NAC$, the segments $BN$ and $NC$ are equal. Thus, we have $[BNC]=\frac{1}{2}BN^2\sin BNC=\frac{1}{2}BN^2\sin A$ since cyclic quadrilateral $ABNC$ gives $\angle BNC=180-\angle A$. Thus, we are to prove that

$\frac{1}{2}BN^2\sin A=[KNC]+[BMN]$

$\Leftrightarrow \frac{1}{2}BN^2\sin A=\frac{1}{2}CN\cdot CK\sin NCA+\frac{1}{2}BN\cdot BM\sin NBA$

$\Leftrightarrow BN\sin A=CK\sin NCA+BM\sin NBA$

From the fact that $\angle BNC=180-\angle A$ and that $BNC$ is iscoceles, we find that $\angle NBC=\angle NCB=\frac{1}{2}A$. So, we have $BN\cos\frac{1}{2}A=\frac{1}{2}BC\Rightarrow BN=\frac{BC}{2\cos \frac{1}{2}A}$. So we are to prove that

$\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA$

$\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)$

$\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)$

We have $\sin C=\frac{KL}{CL}$,$\cos C=\frac{CK}{CL}$, $\cot\frac{1}{2}A=\frac{AK}{KL}=\frac{AM}{LM}$, $\sin B=\frac{LM}{BL}$,$\cos B=\frac{BM}{ML}$, and so we are to prove that

$BC=CK(\frac{KL}{CL}\frac{AK}{KL})+\frac{CK}{CL})+BM(\frac{LM}{BL}\frac{AM}{LM}+\frac{BM}{ML})$

$\Leftrightarrow BC=CK(\frac{AK}{CL}+\frac{CK}{CL})+BM(\frac{AM}{BL}+\frac{BM}{ML})$

$\Leftrightarrow BC=\frac{CK\cdot AC}{CL}+\frac{BM\cdot AB}{BL}$

$\Leftrightarrow BC=AC\cos C+AB\cos B$

We shall show that this is true: Let the altitude from $A$ touch $BC$ at $A^\prime$. Then it is obvious that $AC\cos C=CA^\prime$ and $AB\cos B=A^\prime B$ and thus $AC\cos C+AB\cos B=BC$.

Thus we have proven that $[AKNM]=[ABC]$.

See also

1987 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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