Difference between revisions of "1993 AHSME Problems/Problem 25"
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== Solution == | == Solution == | ||
<math>\fbox{E}</math> | <math>\fbox{E}</math> | ||
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+ | Take the "obvious" equilateral triangle <math>OAP</math>, where <math>O</math> is the vertex, <math>A</math> is on the upper ray, and <math>P</math> is our central point. Slide <math>A</math> down on the top ray to point <math>A'</math>, and slide <math>O</math> down an equal distance on the bottom ray to point <math>O'</math>. | ||
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+ | Now observe <math>\triangle AA'P</math> and <math>\triangle OO'P</math>. We have <math>m\angle A = 60^\circ</math> and <math>m \angle O = 60^\circ</math>, therefore <math>\angle A \cong \angle O</math>. By our construction of moving the points the same distance, we have <math>AA' = OO'</math>. Also, <math>AP = OP</math> by the original equilateral triangle. Therefore, by SAS congruence, <math>\triangle AA'P \cong \triangle OO'P</math>. | ||
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+ | Now, look at <math>\triangle A'PO'</math>. We have <math>PA' = PO'</math> from the above congruence. We also have the included angle <math>\angle A'PO'</math> is <math>60^\circ</math>. To prove that, start with the <math>60^\circ</math> angle <math>APO</math>, subtract the angle <math>APA'</math>, and add the congruent angle <math>OPO'</math>. | ||
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+ | Since <math>\triangle A'PO'</math> is an isosceles triangle with vertex of <math>60^\circ</math>, it is equilateral. | ||
== See also == | == See also == |
Latest revision as of 18:13, 29 July 2019
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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