2001 IMO Problems/Problem 5
Contents
Problem
is a triangle. 
lies on 
and 
bisects angle 
. 
lies on 
and 
bisects angle 
. Angle is 
. 
. Find all possible values for angle .
Solution1
![[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)*A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (0,0.1); dot("$B$", B, NW); dot("$Y$", Y, NE); dot("$D$", D, W); dot("$E$", E, E); dot("$A$",A,N); dot("$X$",X,S); label("$C$",E+(0,-0.1),E); draw(A--D--E--cycle); draw(B--Y); draw(B--E); // draw(B--Xx--E,dashed); // draw(Y--Xx, dashed); draw(A--X--D, dashed); [/asy]](http://latex.artofproblemsolving.com/a/c/c/acc798c0223d06f7394c721e34c44a6ad158dc27.png)
Let 
be on extension of 
and 
. Let 
be on 
and 
, then ![\[AD=AB+BD=AB+BX=AY+YB=AE\]](http://latex.artofproblemsolving.com/c/6/5/c658f97590791eec6fb336862428f42d98725e0a.png)
Since 
, 
is equilateral. Let 
, then, ![\[\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x\]](http://latex.artofproblemsolving.com/5/b/a/5ba07e9033052e4af0c846f3c968bf5bee886eee.png)
We claim that must be on 
, i.e., 
. If is not on , then 
, which leads to 
, and 
is equilateral, which is not possible.
With that, we have, in 
, 
, 
, and 
.
Solution by 
.
Solution 2
Refer to the image in Solution 1 just rename Point X as P and Point Y as Q And no need of construction
\begin{align*} \text{Set: } & \angle ABQ = \angle QBC = x, \quad \angle QCB = 120^\circ - 2x. \\ \text{Observe: } & \angle AQB = 120^\circ - x, \quad \angle APB = 150^\circ - 2x. \\ \text{Using the Law of Sines, we get: } & \\ & AQ = AB \cdot \frac{\sin x^\circ}{\sin(120^\circ - x)}, \\ & BP = AB \cdot \frac{\sin 30^\circ}{\sin(150^\circ - 2x)}, \\ & QB = AB \cdot \frac{\sin 60^\circ}{\sin(120^\circ - x)}. \\ \text{So, the relation } AB + BP &= AQ + AB \text{ is the same as saying} \\ & 1 + \frac{\sin 30^\circ}{\sin(150^\circ - 2x)} = \frac{\sin x + \sin 60^\circ}{\sin(120^\circ - x)}. \\ \text{We have } & \sin x + \sin 60^\circ = 2 \sin\left(\frac{1}{2}(x + 60^\circ)\right) \cos\left(\frac{1}{2}(x - 60^\circ)\right). \\ \text{Also, } & \sin(120^\circ - x) = \sin(x + 60^\circ) \quad \text{and} \\ & \sin(x + 60^\circ) = 2 \sin\left(\frac{1}{2}(x + 60^\circ)\right) \cos\left(\frac{1}{2}(x + 60^\circ)\right). \\ \text{So, } & \frac{\sin x + \sin 60^\circ}{\sin(120^\circ - x)} = \frac{\cos\left(\frac{1}{2}x - 30^\circ\right)}{\cos\left(\frac{1}{2}x + 30^\circ\right)}. \\ \text{Let } & \frac{1}{2}x = t. \\ \text{Then } & \frac{\cos(t - 30^\circ)}{\cos(t + 30^\circ)} - 1 = \frac{\cos(t - 30^\circ) - \cos(t + 30^\circ)}{\cos(t + 30^\circ)} = \frac{2 \sin(30^\circ) \sin(t)}{\cos(t + 30^\circ)}. \\ \text{Hence, the problem is just} & \frac{\sin(30^\circ)}{\sin(150^\circ - 4t)} = \frac{\sin(t)}{\cos(t + 30^\circ)} \\ \Rightarrow & \cos(t + 30^\circ) = 2 \sin(t) \sin(150^\circ - 4t) \\ & = \cos(5t - 150^\circ) - \cos(150^\circ - 3t). \\ \text{Now, } & \cos(t + 30^\circ) + \cos(5t + 30^\circ) = \cos(3t + 30^\circ). \\ \text{Because } & \cos(A + B) + \cos(A - B) = 2\cos A \cos B, \\ \text{we get } & \cos(t + 30^\circ) + \cos(5t + 30^\circ) = 2 \cos(3t + 30^\circ) \cos(2t). \\ \Rightarrow & (2 \cos(2t) - 1)(\cos(3t + 30^\circ)) = 0. \\ \text{This gives } & t \text{ to be } 20^\circ \text{ or } 30^\circ. \\ \text{Recall that } & t = \frac{1}{2}x = \frac{1}{4}\angle ABC. \\ \text{Here we can see } & \angle ABC \neq 120^\circ \text{ because of the angle sum property.} \\ \therefore & \angle B = 80^\circ, \angle A = 60^\circ, \text{ and } \angle C = 40^\circ. \end{align*}
~Lakshya Pamecha
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
| 2001 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
