Difference between revisions of "2004 AMC 12A Problems/Problem 18"
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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}} | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}} | ||
− | ==Problem== | + | == Problem == |
[[Square]] <math>ABCD</math> has side length <math>2</math>. A [[semicircle]] with [[diameter]] <math>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\overline{AD}</math> at <math>E</math>. What is the length of <math>\overline{CE}</math>? | [[Square]] <math>ABCD</math> has side length <math>2</math>. A [[semicircle]] with [[diameter]] <math>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\overline{AD}</math> at <math>E</math>. What is the length of <math>\overline{CE}</math>? | ||
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<math> \mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5} </math> | <math> \mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5} </math> | ||
− | + | == Solutions == | |
− | == Solution 1 == | + | === Solution 1 === |
− | |||
<asy> | <asy> | ||
size(150); | size(150); | ||
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Hence <math>CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}</math>. | Hence <math>CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}</math>. | ||
+ | === 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 Tangent Theorem. Notice that <math>\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF</math>. Thus, <math>\angle EGF=\angle FCG</math> and <math>\tan EGF=\tan FCG=\frac{1}{2}</math>. Solving <math>EF=\frac{1}{2}</math>. Adding, the answer is <math>\frac{5}{2}</math>. | ||
− | = | + | === Solution 3 === |
− | |||
− | |||
− | == Solution 3 == | ||
− | |||
[[Image:2004_AMC12A-18.png]] | [[Image:2004_AMC12A-18.png]] | ||
− | Clearly, <math>EA = EF = BG</math>. Thus, the sides of [[right triangle]] <math>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <math>CE = 5 | + | Clearly, <math>EA = EF = BG</math>. Thus, the sides of [[right triangle]] <math>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <math>CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}</math>. |
− | == Solution 4 == | + | === Solution 4 === |
<asy> | <asy> | ||
size(150); | size(150); | ||
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Let us call the midpoint of side <math>AB</math>, point <math>G</math>. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides <math>GB, BC, GC</math>. We get <math>GC=\sqrt{5}</math>. We then know that <math>CF=2</math> by Pythagorean theorem. Then by connecting <math>EG</math>, we get similar triangles <math>EFG</math> and <math>GFC</math>. Solving the ratios, we get <math>x=\frac{1}{2}</math>, so the answer is <math> \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}</math>. | Let us call the midpoint of side <math>AB</math>, point <math>G</math>. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides <math>GB, BC, GC</math>. We get <math>GC=\sqrt{5}</math>. We then know that <math>CF=2</math> by Pythagorean theorem. Then by connecting <math>EG</math>, we get similar triangles <math>EFG</math> and <math>GFC</math>. Solving the ratios, we get <math>x=\frac{1}{2}</math>, so the answer is <math> \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}</math>. | ||
− | |||
− | |||
− | <cmath>DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = | + | Alternatively, we could apply the Pythagorean theorem on triangle <math>CDE</math> to get the equation <cmath>(2-x)^2+2^2=(x+2)^2</cmath> which would give us that <math>x=\frac{1}{2}.</math> Adding up <math>CF</math> and <math>EF,</math> we get <math>2+\frac{1}{2}=\boxed{\mathrm{(D)}\ \frac{5}{2}}</math> again. Note that we know <math>DE=2-x</math> because <math> \triangle EFG\cong \triangle EAG</math> by <math>\text{HL.}</math> |
+ | |||
+ | ~Alternate solution by Tinsel | ||
+ | |||
+ | === Solution 5 === | ||
+ | Using the diagram as drawn in Solution 4, let the total area of square <math>ABCD</math> be divided into the triangles <math>DCE</math>, <math>EAG</math>, <math>CGB</math>, and <math>EGC</math>. Let x be the length of AE. Thus, the area of each triangle can be determined as follows: | ||
+ | |||
+ | <cmath>DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 2-x</cmath> | ||
<cmath>EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}</cmath> | <cmath>EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}</cmath> | ||
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<cmath>CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1</cmath> | <cmath>CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1</cmath> | ||
− | <cmath>EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{ | + | <cmath>EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}</cmath> (the length of CE is calculated with the [[Pythagorean Theorem]], lines GE and |
CE are perpendicular by definition of tangent) | CE are perpendicular by definition of tangent) | ||
− | Adding up the areas and equating to the area of the total square (2 | + | Adding up the areas and equating to the area of the total square <cmath>(2 \cdot 2=4)</cmath>, we get |
+ | |||
+ | <cmath>x = \frac{1}{2}</cmath> | ||
− | <cmath> | + | So, <cmath>CE = 2 + \frac{1}{2} = 5/2</cmath>. |
− | + | ~Typo Fix by doulai1 | |
− | + | === Video Solution === | |
− | + | https://youtu.be/pM0zICtH6Lg | |
− | |||
− | |||
− | + | Education, the Study of Everything | |
− | |||
− | == See | + | == See Also == |
− | |||
{{AMC12 box|year=2004|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2004|ab=A|num-b=17|num-a=19}} | ||
+ | |||
{{AMC10 box|year=2004|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2004|ab=A|num-b=21|num-a=23}} | ||
− | |||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:30, 27 June 2024
- The following problem is from both the 2004 AMC 12A #18 and 2004 AMC 10A #22, so both problems redirect to this page.
Contents
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
?
Solutions
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 Tangent Theorem. Notice that
. Thus,
and
. Solving
. 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 4
Let us call the midpoint of side , point
. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides
. We get
. We then know that
by Pythagorean theorem. Then by connecting
, we get similar triangles
and
. Solving the ratios, we get
, so the answer is
.
Alternatively, we could apply the Pythagorean theorem on triangle to get the equation
which would give us that
Adding up
and
we get
again. Note that we know
because
by
~Alternate solution by Tinsel
Solution 5
Using the diagram as drawn in Solution 4, let the total area of square be divided into the triangles
,
,
, and
. Let x be the length of AE. Thus, the area of each triangle can be determined as follows:
(the length of CE is calculated with the Pythagorean Theorem, lines GE and
CE are perpendicular by definition of tangent)
Adding up the areas and equating to the area of the total square , we get
So, .
~Typo Fix by doulai1
Video Solution
Education, the Study of Everything
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.