2013 UMO Problems/Problem 4

Revision as of 21:16, 16 February 2020 by Helloitsaaryan (talk | contribs)

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.


See Also

2013 UMO (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All UMO Problems and Solutions