Difference between revisions of "1983 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
A machine shop cutting tool is in the shape of a notched [[circle]], as shown. The [[radius]] of the circle is <math>\sqrt{50}</math> cm, the [[length]] of <math>AB</math> is 6 cm, and that of <math>BC</math> is 2 cm. The [[angle]] <math>ABC</math> is a [[right angle]]. Find the [[square]] of the distance (in centimeters) from <math>B</math> to the center of the circle. | A machine shop cutting tool is in the shape of a notched [[circle]], as shown. The [[radius]] of the circle is <math>\sqrt{50}</math> cm, the [[length]] of <math>AB</math> is 6 cm, and that of <math>BC</math> is 2 cm. The [[angle]] <math>ABC</math> is a [[right angle]]. Find the [[square]] of the distance (in centimeters) from <math>B</math> to the center of the circle. | ||
− | + | <center><asy> | |
+ | size(150); defaultpen(linewidth(0.6)+fontsize(11)); | ||
+ | real r=10; | ||
+ | pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; | ||
+ | path P=circle(O,r); | ||
+ | C=intersectionpoint(B--(B.x+r,B.y),P); | ||
+ | draw(P); | ||
+ | draw(C--B--O--A--B); | ||
+ | dot(O); dot(A); dot(B); dot(C); | ||
+ | label("$O$",O,SW); | ||
+ | label("$A$",A,NE); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,SE); | ||
+ | </asy></center> | ||
== Solution == | == Solution == | ||
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>. | Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]]. Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>. | ||
− | + | <center><asy> | |
+ | size(150); defaultpen(linewidth(0.6)+fontsize(11)); | ||
+ | real r=10; | ||
+ | pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; | ||
+ | pair D=(A.x,0),F=(0,B.y); | ||
+ | path P=circle(O,r); | ||
+ | C=intersectionpoint(B--(B.x+r,B.y),P); | ||
+ | draw(P); | ||
+ | draw(C--B--O--A--B); | ||
+ | draw(D--O--F--B,dashed); | ||
+ | dot(O); dot(A); dot(B); dot(C); | ||
+ | label("$O$",O,SW); | ||
+ | label("$A$",A,NE); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",F,SW); | ||
+ | </asy></center><!-- Asymptote replacement for Image:AIME_83_-4_Modified.JPG by bpms --> | ||
Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</math>. | Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</math>. | ||
− | Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = 026</math>. | + | Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = \boxed{026}</math>. |
== See also == | == See also == |
Revision as of 11:01, 25 April 2008
Problem
A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is cm, the length of is 6 cm, and that of is 2 cm. The angle is a right angle. Find the square of the distance (in centimeters) from to the center of the circle.
Solution
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 , resulting in an answer of .
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 |