AoPSWiki

Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.

Personal tools

Search

Toolbox

1987 IMO Problems/Problem 2

From AoPSWiki

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)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1987_IMO_Problems/Problem_2"

Category: Olympiad Geometry Problems

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us