Difference between revisions of "2004 AMC 12A Problems/Problem 18"
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label("$x$",(A+E)/2,(-1, 0)); | label("$x$",(A+E)/2,(-1, 0)); | ||
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Revision as of 14:22, 25 July 2015
- 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
Clearly, . Thus, the sides of right triangle
are in arithmetic progression. Thus it is similar to the triangle
and since
,
.
Solution 3
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label( (Error making remote request. Unknown error_msg)
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.