Difference between revisions of "2013 UMO Problems/Problem 4"
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== Solution == | == 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. | |
Revision as of 20:16, 16 February 2020
Problem
Given line 
and distinct points 
and 
on line , draw lines 
and 
through point , with angles 
and 
as marked in the figure. Also, draw line segment 
at an angle of 
from line such that it intersects line at 
. Establish necessary and sufficient conditions on , , and such that a triangle can be drawn with one of its sides as with lines , , and 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]](http://latex.artofproblemsolving.com/e/7/9/e797a0b083710b51f141ae818ad1ea7a15c0d6b0.png)
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 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All UMO Problems and Solutions | ||
