Difference between revisions of "1993 AHSME Problems/Problem 25"
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== Problem == | == Problem == | ||
<asy> | <asy> | ||
− | draw((0,0)--(1,sqrt(3)),black+linewidth(.75)); | + | draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow); |
− | draw((0,0)--(1,-sqrt(3)),black+linewidth(.75)); | + | draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow); |
draw((0,0)--(1,0),dashed+black+linewidth(.75)); | draw((0,0)--(1,0),dashed+black+linewidth(.75)); | ||
dot((1,0)); | dot((1,0)); | ||
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== Solution == | == Solution == | ||
<math>\fbox{E}</math> | <math>\fbox{E}</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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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