Difference between revisions of "2001 IMO Problems/Problem 5"
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==Problem== | ==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 | + | <math>ABC</math> is a [[triangle]]. <math>X</math> lies on <math>BC</math> and <math>AX</math> bisects [[angle]] <math>A</math>. <math>Y</math> lies on <math>CA</math> and <math>BY</math> bisects angle <math>B</math>. Angle <math>A</math> is <math>60^{\circ}</math>. <math>AB + BX = AY + YB</math>. Find all possible values for angle <math>B</math>. |
− | ==Solution== | + | ==Solution1 == |
− | {{ | + | <center><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></center> | ||
+ | Let <math>D</math> be on extension of <math>AB</math> and <math>BD=BX</math>. Let <math>E</math> be on <math>YC</math> and <math>YE=YB</math>, then <cmath>AD=AB+BD=AB+BX=AY+YB=AE</cmath> | ||
+ | Since <math>A=60</math>, <math>\triangle{ADE}</math> is equilateral. Let <math>\angle{ABY}=x</math>, then, <cmath>\angle{YBX}=\angle{BDX}=\angle{BXD}=\angle{YEX}=x</cmath> | ||
+ | We claim that <math>X</math> must be on <math>BE</math>, i.e., <math>C=E</math>. If <math>X</math> is not on <math>BE</math>, then <math>\angle{EBX}=\angle{YBX}-\angle{YBE}=\angle{YEX}-\angle{YEB}=\angle{BEX}</math>, which leads to <math>BX=EX=DX</math>, and <math>\triangle{BDX}</math> is equilateral, which is not possible. | ||
+ | With that, we have, in <math>\triangle{ABE}</math>, <math>60+2x+x=180</math>, <math>x=40</math>, and <math>\angle{ABE}=80</math>. | ||
+ | |||
+ | Solution by <math>Mathdummy</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Refer to the image in Solution 1 without any construction | ||
+ | \begin{align*} | ||
+ | \text{Set: } & \angle ABY = \angle YBC = x, \quad \angle YCB = 120^\circ - 2x. \ | ||
+ | \text{Observe: } & \angle AYB = 120^\circ - x, \quad \angle AXB = 150^\circ - 2x. \ | ||
+ | \text{Using the Law of Sines, we get: } & \ | ||
+ | & AY = AB \cdot \frac{\sin x^\circ}{\sin(120^\circ - x)}, \ | ||
+ | & BX = AB \cdot \frac{\sin 30^\circ}{\sin(150^\circ - 2x)}, \ | ||
+ | & YB = AB \cdot \frac{\sin 60^\circ}{\sin(120^\circ - x)}. \ | ||
+ | \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)}. \ | ||
+ | \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}} | ||
==See also== | ==See also== | ||
+ | {{IMO box|num-b=4|num-a=6|year=2001}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 14:01, 15 May 2024
Contents
[hide]Problem
is a triangle. lies on and bisects angle . lies on and bisects angle . Angle is . . Find all possible values for angle .
Solution1
Let be on extension of and . Let be on and , then Since , is equilateral. Let , then, 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 without any construction
~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 |