Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> | <math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> | ||
− | + | == Solution 2 == | |
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− | == Solution 2== | ||
<math>APMD</math> obviously forms a kite. Let the intersection of the diagonals be <math>E</math>. <math>AE+EM=AM=2\sqrt{5}</math> Let <math>AE=x</math>. Then, <math>EM=2\sqrt{5}-x</math>. | <math>APMD</math> obviously forms a kite. Let the intersection of the diagonals be <math>E</math>. <math>AE+EM=AM=2\sqrt{5}</math> Let <math>AE=x</math>. Then, <math>EM=2\sqrt{5}-x</math>. | ||
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− | Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}</math>. <math>\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5} \implies \boxed{\textbf{( | + | Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}</math>. <math>\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}</math> |
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~BJHHar | ~BJHHar | ||
− | ==Solution 3== | + | == Solution 3 == |
Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>. | Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>. | ||
− | ==Solution 4== | + | == Solution 4 == |
As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem. | As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem. | ||
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Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math> | Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math> | ||
− | ==Solution 5== | + | == Solution 5 == |
Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>. | Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>. | ||
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==Solution 6(LoC)== | ==Solution 6(LoC)== | ||
We use the Law of Cosines: | We use the Law of Cosines: | ||
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<math>2 + 2*\frac{3}{5} = \frac{16}{5}</math> | <math>2 + 2*\frac{3}{5} = \frac{16}{5}</math> | ||
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+ | == Solution 7 == | ||
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+ | Let <math>H</math> be the foot of the perpendicular from <math>P</math> to <math>CD</math>, and let <math>HD = x</math>. Then we have <math>HC = 4-x</math>, and <math>PH = 4 - \sqrt{16 - x^2}</math>. Since <math>\triangle DHP \sim \triangle PHC</math>, we have <math>HP^2 = DH \cdot HC</math>, or <math>-x^2 + 4x = 16 - 8\sqrt{16-x^2}</math>. Solving gives <math>x = \frac{16}{5}</math>. | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm, 8cm); | ||
+ | pair A,B,C,D,M,P,Q,R; | ||
+ | D = (0,0); | ||
+ | C = (10,0); | ||
+ | B = (10,10); | ||
+ | A = (0,10); | ||
+ | M = (5,0); | ||
+ | P = (8,4); | ||
+ | Q = (D+P)/2; | ||
+ | R = (0,4); | ||
+ | dot(M); | ||
+ | dot(P); | ||
+ | draw(A--B--C--D--cycle,linewidth(0.7)); | ||
+ | draw((5,5)..D--C..cycle,linewidth(0.7)); | ||
+ | draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7)); | ||
+ | draw(A--M,linewidth(0.7)); | ||
+ | draw(A--P,linewidth(0.7)); | ||
+ | draw(D--P,linewidth(0.7)); | ||
+ | draw(R--P,linewidth(0.7)); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,SW); | ||
+ | label("$M$",M,S); | ||
+ | label("$P$",P,N); | ||
+ | label("$Q$",Q,W); | ||
+ | label("$R$",R,W); | ||
+ | |||
+ | draw(rightanglemark(M, Q, P), linewidth(.5)); | ||
+ | </asy> | ||
+ | |||
+ | Draw <math>AM</math>, <math>DP</math>, and <math>PR</math>. <math>PR</math> is parallel with <math>CD</math> | ||
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+ | <math>[AMD] = \frac12 \cdot AD \cdot DM = 4</math>, <math>AM = \sqrt{AD^2 + DM^2} = 2 \sqrt{5}</math> | ||
+ | |||
+ | <math>\triangle ADQ \sim \triangle AMD</math> by <math>AA</math>, <math>[ADQ] = [AMD] \cdot \left( \frac{AD}{AM} \right) ^2 = 4 \cdot \left( \frac{2 \sqrt{5}}{5} \right)^2 = \frac{16}{5}</math> | ||
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+ | <math>\triangle ADQ \cong \triangle APQ</math>, <math>[APD] = 2 \cdot [ADQ] = 2 \cdot \frac{16}{5} = \frac{32}{5}</math> | ||
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+ | <math>PR = \frac{2 \cdot [APD]}{AD} = \frac{2 \cdot \frac{32}{5}}{4} = \boxed{\textbf{(B) } \frac{16}{5} }</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:43, 19 September 2024
Contents
Problem
Square has sides of length , and is the midpoint of . A circle with radius and center intersects a circle with radius and center at points and . What is the distance from to ?
Solution 2
obviously forms a kite. Let the intersection of the diagonals be . Let . Then, .
By Pythagorean Theorem, . Thus, . Simplifying, . By Pythagoras again, . Then, the area of is .
Using instead as the base, we can drop a altitude from P. . . Thus, the horizontal distance is
~BJHHar
Solution 3
Note that is merely a reflection of over . Call the intersection of and . Drop perpendiculars from and to , and denote their respective points of intersection by and . We then have , with a scale factor of 2. Thus, we can find and double it to get our answer. With some analytical geometry, we find that , implying that .
Solution 4
As in Solution 2, draw in and and denote their intersection point . Next, drop a perpendicular from to and denote the foot as . as they are both radii and similarly so is a kite and by a well-known theorem.
Pythagorean theorem gives us . Clearly by angle-angle and by Hypotenuse Leg. Manipulating similar triangles gives us
Solution 5
Using the double-angle formula for sine, what we need to find is .
Solution 6(LoC)
We use the Law of Cosines:
Solution 7
Let be the foot of the perpendicular from to , and let . Then we have , and . Since , we have , or . Solving gives .
Solution 8
Draw , , and . is parallel with
,
by ,
,
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.