1993 AHSME Problems/Problem 25
Problem
Let be the set of points on the rays forming the sides of a angle, and let be a fixed point inside the angle on the angle bisector. Consider all distinct equilateral triangles with and in . (Points and may be on the same ray, and switching the names of and does not create a distinct triangle.) There are
Solution
Take the "obvious" equilateral triangle , where is the vertex, is on the upper ray, and is our central point. Slide down on the top ray to point , and slide down an equal distance on the bottom ray to point .
Now observe and . We have and , therefore . By our construction of moving the points the same distance, we have . Also, by the original equilateral triangle. Therefore, by SAS congruence, .
Now, look at . We have from the above congruence. We also have the included angle is . To prove that, start with the angle , subtract the angle , and add the congruent angle .
Since is an isosceles triangle with vertex of , it is equilateral.
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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