Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math> | The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math> | ||
− | ==Solution 2== | + | == 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>. By Pythagorean Theorem, <math>DE^2=4^2-AE^2=2^2-EM^2</math>. Thus, <math>16-x^2=4-(2\sqrt{5}-x)^2</math>. Simplifying, <math>x=\frac{8}{\sqrt{5}}</math>. By Pythagoras again, <math>DE=\frac{4}{\sqrt{5}}</math>. Then, the area of <math>ADP</math> is <math>DE\cdot AE=\frac{32}{5}</math>. Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}\implies\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5}\implies\boxed{\textbf{(E)\frac{16}{5}}</math> | ||
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+ | ==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 | + | ==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 | + | ==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>. | ||
Revision as of 00:10, 28 September 2019
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 1
Let be the origin. is the point and is the point . We are given the radius of the quarter circle and semicircle as and , respectively, so their equations, respectively, are:
Subtract the second equation from the first:
Then substitute:
Thus and making and .
The first value of is obviously referring to the x-coordinate of the point where the circles intersect at the origin, , so the second value must be referring to the x coordinate of . Since is the y-axis, the distance to it from is the same as the x-value of the coordinate of , so the distance from to is
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 $\frac{16}{5}\implies\boxed{\textbf{(E)\frac{16}{5}}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)
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 .
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 |
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