Difference between revisions of "2004 AMC 12A Problems/Problem 18"
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== Solution 2 == | == Solution 2 == | ||
− | Call the point of tangency point <math>F</math> and the midpoint of <math>AB</math> as <math>G</math>. <math>CF=2</math> by the pythagorean theorem. Notice that <math>\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF. Thus, EF=\cot \theta | + | Call the point of tangency point <math>F</math> and the midpoint of <math>AB</math> as <math>G</math>. <math>CF=2</math> by the pythagorean theorem. Notice that <math>\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF</math>. Thus, <math>EF=\cot \theta=\frac{1}{2}</math>. Adding, the answer is <math>\frac{5}{2}</math>. |
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== Solution 3 == | == Solution 3 == |
Revision as of 14:08, 8 February 2016
- The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.
Problem
Square has side length
. A semicircle with diameter
is constructed inside the square, and the tangent to the semicircle from
intersects side
at
. What is the length of
?
Solution 1
Let the point of tangency be
. By the Two Tangent Theorem
and
. Thus
. The Pythagorean Theorem on
yields
Hence .
Solution 2
Call the point of tangency point and the midpoint of
as
.
by the pythagorean theorem. Notice that
. Thus,
. Adding, the answer is
.
Solution 3
Clearly, . Thus, the sides of right triangle
are in arithmetic progression. Thus it is similar to the triangle
and since
,
.
Solution 3
TBE
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
2004 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.