Difference between revisions of "2002 AMC 12A Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | Clearly <math>BC=5</math> and <math>AC=5\sqrt{3}</math>. Choose a <math>P'</math> and get a corresponding <math>D'</math> such that <math>BD'= 5\sqrt{2}</math> and <math>CD'=5</math>. For <math> BD > 5\sqrt2 </math> we need <math>CD>5</math>, creating an | + | Clearly <math>BC=5</math> and <math>AC=5\sqrt{3}</math>. Choose a <math>P'</math> and get a corresponding <math>D'</math> such that <math>BD'= 5\sqrt{2}</math> and <math>CD'=5</math>. For <math> BD > 5\sqrt2 </math> we need <math>CD>5</math>, creating an isosceles right triangle with hyptenuse <math>5\sqrt {2}</math> . Thus the point <math>P</math> may only lie in the triangle <math>ABD'</math>. The probability of it doing so is the ratio of areas of <math>ABD'</math> to <math>ABC</math>, or equivalently, the ratio of <math>AD'</math> to <math>AC</math> because the triangles have identical altitudes when taking <math>AD'</math> and <math>AC</math> as bases. This ratio is equal to <math>\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}</math>. Thus the answer is <math>\boxed{C}</math>. |
==See Also== | ==See Also== |
Revision as of 23:03, 3 April 2019
Problem
Triangle is a right triangle with
as its right angle,
, and
. Let
be randomly chosen inside
, and extend
to meet
at
. What is the probability that
?
Solution
Clearly and
. Choose a
and get a corresponding
such that
and
. For
we need
, creating an isosceles right triangle with hyptenuse
. Thus the point
may only lie in the triangle
. The probability of it doing so is the ratio of areas of
to
, or equivalently, the ratio of
to
because the triangles have identical altitudes when taking
and
as bases. This ratio is equal to
. Thus the answer is
.
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 | |
All AMC 12 Problems and Solutions |
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