Difference between revisions of "2001 IMO Problems/Problem 5"

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(Solution 2)
 
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==Solution 2==
 
==Solution 2==
<asy>
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Refer to the image in Solution 1 without any construction
import cse5;
 
import graph;
 
import olympiad;
 
dotfactor = 3;
 
unitsize(1.5inch);
 
 
 
pair A = (0,sqrt(3)), E=(1,0);
 
pair Bb = rotate(40,E)*A;
 
pair B = extension(A,E,E,Bb);
 
pair H = foot(A,E,E);
 
pair Q = extension(A,H,B,E);
 
pair Yy = bisectorpoint(A,B,E);
 
pair P = extension(A,E,B,Yy);
 
pair C = E - (0,0.1);
 
 
 
 
 
dot("$B$", B, NW); dot("$P$", P, NE);
 
dot("$E$", E, E);
 
dot("$A$",A,N); dot("$Q$",Q,S);
 
label("$C$",E+(0,-0.1),E);
 
 
 
draw(A--E--C--cycle);
 
draw(B--P);
 
draw(B--E);
 
draw(A--Q--E, dashed);
 
</asy>
 
 
 
 
 
 
\begin{align*}
 
\begin{align*}
\text{Set: } & \angle ABQ = \angle QBC = x, \quad \angle QCB = 120^\circ - 2x. \
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\text{Set: } & \angle ABY = \angle YBC = x, \quad \angle YCB = 120^\circ - 2x. \
\text{Observe: } & \angle AQB = 120^\circ - x, \quad \angle APB = 150^\circ - 2x. \
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\text{Observe: } & \angle AYB = 120^\circ - x, \quad \angle AXB = 150^\circ - 2x. \
 
\text{Using the Law of Sines, we get: } & \
 
\text{Using the Law of Sines, we get: } & \
& AQ = AB \cdot \frac{\sin x^\circ}{\sin(120^\circ - x)}, \
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& AY = AB \cdot \frac{\sin x^\circ}{\sin(120^\circ - x)}, \
& BP = AB \cdot \frac{\sin 30^\circ}{\sin(150^\circ - 2x)}, \
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& BX = AB \cdot \frac{\sin 30^\circ}{\sin(150^\circ - 2x)}, \
& QB = AB \cdot \frac{\sin 60^\circ}{\sin(120^\circ - x)}. \
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& YB = AB \cdot \frac{\sin 60^\circ}{\sin(120^\circ - x)}. \
\text{So, the relation } AB + BP &= AQ + AB \text{ is the same as saying} \
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\text{So, the relation } AB + BX &= AY + 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)}. \
 
& 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{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). \

Latest revision as of 14:01, 15 May 2024

Problem

$ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$. $Y$ lies on $CA$ and $BY$ bisects angle $B$. Angle $A$ is $60^{\circ}$. $AB + BX = AY + YB$. Find all possible values for angle $B$.

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]

Let $D$ be on extension of $AB$ and $BD=BX$. Let $E$ be on $YC$ and $YE=YB$, then \[AD=AB+BD=AB+BX=AY+YB=AE\] Since $A=60$, $\triangle{ADE}$ is equilateral. Let $\angle{ABY}=x$, then, \[\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x\] We claim that $X$ must be on $BE$, i.e., $C=E$. If $X$ is not on $BE$, then $\angle{EBX}=\angle{YBX}-\angle{YBE}=\angle{YEX}-\angle{YEB}=\angle{BEX}$, which leads to $BX=EX=DX$, and $\triangle{BDX}$ is equilateral, which is not possible. With that, we have, in $\triangle{ABE}$, $60+2x+x=180$, $x=40$, and $\angle{ABE}=80$.

Solution by $Mathdummy$.

Solution 2

Refer to the image in Solution 1 without any construction Set: ABY=YBC=x,YCB=1202x.Observe: AYB=120x,AXB=1502x.Using the Law of Sines, we get: AY=ABsinxsin(120x),BX=ABsin30sin(1502x),YB=ABsin60sin(120x).So, the relation AB+BX=AY+AB is the same as saying1+sin30sin(1502x)=sinx+sin60sin(120x).We have sinx+sin60=2sin(12(x+60))cos(12(x60)).Also, sin(120x)=sin(x+60)andsin(x+60)=2sin(12(x+60))cos(12(x+60)).So, sinx+sin60sin(120x)=cos(12x30)cos(12x+30).Let 12x=t.Then cos(t30)cos(t+30)1=cos(t30)cos(t+30)cos(t+30)=2sin(30)sin(t)cos(t+30).Hence, the problem is justsin(30)sin(1504t)=sin(t)cos(t+30)cos(t+30)=2sin(t)sin(1504t)=cos(5t150)cos(1503t).Now, cos(t+30)+cos(5t+30)=cos(3t+30).Because cos(A+B)+cos(AB)=2cosAcosB,we get cos(t+30)+cos(5t+30)=2cos(3t+30)cos(2t).(2cos(2t)1)(cos(3t+30))=0.This gives t to be 20 or 30.Recall that t=12x=14ABC.Here we can see ABC120 because of the angle sum property.B=80,A=60, and C=40.

~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