Difference between revisions of "1970 AHSME Problems/Problem 7"
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== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | All answers are proportional to <math>s</math>, so for ease, let <math>s=1</math>. |
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+ | Let <math>ABCD</math> be oriented so that <math>A(0, 0), B(1, 0), C(1, 1), D(0, 1)</math>. | ||
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+ | The circle centered at <math>A</math> with radius <math>1</math> is <math>x^2 + y^2 = 1</math>. The circle centered at <math>B</math> with radius <math>1</math> is <math>(x - 1)^2 + y^2 = 1</math>. Solving each equation for <math>1 - y^2</math> to find the intersection leads to <math>(x - 1)^2 = x^2</math>, which leads to <math>x = \frac{1}{2}</math>. | ||
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+ | Plugging that back in to <math>x^2 + y^2 = 1</math> leads to <math>y^2 = 1 - \frac{1}{4}</math>, or <math>y = \pm \frac{\sqrt{3}}{2}</math>. Since we want the intersection within the square where <math>0 < x, y < 1</math>, we take the positive solution, and the intersection is at <math>X(\frac{1}{2}, \frac{\sqrt{3}}{2}}</math>. | ||
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+ | The distance from <math>X</math> to side <math>CD</math>, which lies along the line <math>y=1</math>, is <math>1 - \frac{\sqrt{3}}{2}</math>, or <math>\frac{2 -\sqrt{3}}{2}</math>. This is answer <math>\fbox{E}</math> when <math>s=1</math>. | ||
== See also == | == See also == |
Revision as of 19:53, 13 July 2019
Problem
Inside square with side , quarter-circle arcs with radii and centers at and are drawn. These arcs intersect at a point inside the square. How far is from the side of ?
Solution
All answers are proportional to , so for ease, let .
Let be oriented so that .
The circle centered at with radius is . The circle centered at with radius is . Solving each equation for to find the intersection leads to , which leads to .
Plugging that back in to leads to , or . Since we want the intersection within the square where , we take the positive solution, and the intersection is at $X(\frac{1}{2}, \frac{\sqrt{3}}{2}}$ (Error making remote request. No response to HTTP request).
The distance from to side , which lies along the line , is , or . This is answer when .
See also
1970 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.