Difference between revisions of "1983 AIME Problems/Problem 4"
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Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath> | Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath> | ||
+ | ===Solution 4=== | ||
+ | We use coordinates. Let the circle have center <math>(0,0)</math> and radius <math>\sqrt{50}</math>; this circle has equation <math>x^2 + y^2 = 50</math>. Let the coordinates of <math>B</math> be <math>(a,b)</math>. We want to find <math>a^2 + b^2</math>. <math>A</math> and <math>C</math> with coordinates <math>(a,b+6)</math> and <math>(a+2,b)</math>, respectively, both lie on the circle. From this we obtain the system of equations | ||
+ | <math>a^2 + (b+6)^2 = 50</math> | ||
+ | |||
+ | <math>(a+2)^2 + b^2 = 50</math> | ||
+ | |||
+ | Solving, we get <math>a=5</math> and <math>b=-1</math>, so the distance is <math>a^2 + b^2 = \boxed{026}</math>. | ||
== See Also == | == See Also == | ||
{{AIME box|year=1983|num-b=3|num-a=5}} | {{AIME box|year=1983|num-b=3|num-a=5}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 21:42, 30 October 2020
Contents
[hide]Problem
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is cm, the length of is cm and that of is cm. The angle is a right angle. Find the square of the distance (in centimeters) from to the center of the circle.
Solution
Solution 1
Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from to be and let the foot of the perpendicular from to the line be . Let and . We're trying to find .
Applying the Pythagorean Theorem, and .
Thus, , and . We solve this system to get and , such that the answer is .
Solution 2
Drop perpendiculars from to (with foot ), to (with foot ), and to (with foot ). Also, mark the midpoint of .
Then the problem is trivialized. Why?
First notice that by computation, is a isosceles triangle, so . Then, notice that . Therefore, the two blue triangles are congruent, from which we deduce and . As and , we subtract and get . Then the Pythagorean Theorem tells us that .
Solution 3
Draw segment with length , and draw radius such that bisects chord at point . This also means that is perpendicular to . By the Pythagorean Theorem, we get that , and therefore . Also by the Pythagorean theorem, we can find that .
Next, find and . Since , we get By the subtraction formula for , we getFinally, by the Law of Cosines on , we get
Solution 4
We use coordinates. Let the circle have center and radius ; this circle has equation . Let the coordinates of be . We want to find . and with coordinates and , respectively, both lie on the circle. From this we obtain the system of equations
Solving, we get and , so the distance is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |