Difference between revisions of "2004 AMC 12A Problems/Problem 18"
(→Solution 3) |
(→Solution 3) |
||
Line 59: | Line 59: | ||
== Solution 3 == | == Solution 3 == | ||
+ | [[asy] 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("<math>A</math>",A,(-1,-1)); label("<math>B</math>",B,( 1,-1)); label("<math>C</math>",C,( 1, 1)); label("<math>D</math>",D,(-1, 1)); label("<math>E</math>",E,(-1, 0)); label("<math>F</math>",F,( 0, 1)); label("<math>x</math>",(A+E)/2,(-1, 0)); label("<math>x</math>",(E+F)/2,( 0, 1)); label("<math>2</math>",(F+C)/2,( 0, 1)); label("<math>2</math>",(D+C)/2,( 0, 1)); label("<math>2</math>",(B+C)/2,( 1, 0)); label("<math>2-x</math>",(D+E)/2,(-1, 0)); [/asy]] | ||
== See also == | == See also == |
Revision as of 13:16, 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
?
Contents
[hide]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
[[asy] 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,(-1,-1)); label("
",B,( 1,-1)); label("
",C,( 1, 1)); label("
",D,(-1, 1)); label("
",E,(-1, 0)); label("
",F,( 0, 1)); label("
",(A+E)/2,(-1, 0)); label("
",(E+F)/2,( 0, 1)); label("
",(F+C)/2,( 0, 1)); label("
",(D+C)/2,( 0, 1)); label("
",(B+C)/2,( 1, 0)); label("
",(D+E)/2,(-1, 0)); [/asy]]
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.