Difference between revisions of "1954 AHSME Problems/Problem 46"
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− | First we extend the line with <math>A</math> and the line <math>B</math> so that they both meet the line with <math>C</math>, forming an equilateral triangle. Let the vertices of this triangle be <math>D</math>, <math>E</math>, and <math>F</math>. We know it is equilateral because of the angle of <math>60^\circ</math> shown, and because the tangent lines <math>\overline{EF}</math> and <math>\overline{DE}</math> are congruent. <asy> | + | First we extend the line with <math>A</math> and the line with <math>B</math> so that they both meet the line with <math>C</math>, forming an equilateral triangle. Let the vertices of this triangle be <math>D</math>, <math>E</math>, and <math>F</math>. We know it is equilateral because of the angle of <math>60^\circ</math> shown, and because the tangent lines <math>\overline{EF}</math> and <math>\overline{DE}</math> are congruent. <asy> |
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Latest revision as of 18:01, 24 May 2020
Problem 46
In the diagram, if points and
are points of tangency, then
equals:
Solution 1
First we extend the line with and the line with
so that they both meet the line with
, forming an equilateral triangle. Let the vertices of this triangle be
,
, and
. We know it is equilateral because of the angle of
shown, and because the tangent lines
and
are congruent.
We can see, because
,
, and
are points of tangency, that circle
is inscribed in
. The height of an equilateral triangle is exactly
times the radius of a circle inscribed in it. Let the height of
be
. We can see that the radius of the circle equals
. Thus
Subtracting
from
gives us
so our answer is
.