2013 UMO Problems/Problem 4

Problem

Given line $\ell_1$ and distinct points $I$ and $X$ on line $\ell_1$ , draw lines $\ell_2$ and $\ell_3$ through point $I$ , with angles $\alpha$ and $\beta$ as marked in the figure. Also, draw line segment $XY$ at an angle of $\gamma$ from line $\ell_1$ such that it intersects line $\ell_2$ at $Y$ . Establish necessary and sufficient conditions on $\alpha$ , $\beta$ , and $\gamma$ such that a triangle can be drawn with one of its sides as $XY$ with lines $\ell_1$ , $\ell_2$ , and $\ell_3$ as the angle bisectors of that triangle.

$[asy] D((-2,-2)--(2,2),black+linewidth(.75)); D((-1,2)--(1,-2),black+linewidth(.75)); D((-2,1)--(2,-1),black+linewidth(.75)); D((-.7,1.4)--(-1.3,-1.3),black+linewidth(.75)); MP("\ell_1",(1.5,1.6),N); MP("\ell_2",(-1.1,1.5),N); MP("\ell_3",(-1.5,.3),N); MP("\alpha",(.05,.2),N); MP("\beta",(-.35,.13),N); MP("\gamma",(-1.1,-1.1),N); MP("Y",(-.7,1.4),NE); MP("X",(-1.3,-1.3),S); MP("I",(-.1,-.1),S); dot((-1.3,-1.3));dot((-.7,1.4));dot((0,0)); [/asy]$

Solution

As we need an angle bisector, let the triangle XYI be our desired triangle. The bisected angle is clearly \beta. Using the fact that a straight line is 180 degrees, we have \alpha+2*\beta = 180 as our condition.

2013 UMO (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
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2013 UMO Problems/Problem 4

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