Difference between revisions of "1983 AIME Problems/Problem 4"

Revision as of 12:01, 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 $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]$

Solution

Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$ .

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",F,SW); [/asy]$

Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$ .

Thus, $(\sqrt{50})^2 = y^2 + (6-x)^2$ , and $(\sqrt{50})^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , resulting in an answer of $1^2 + 5^2 = \boxed{026}$ .

1983 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1983 AIME Problems/Problem 4"

Revision as of 12:01, 25 April 2008

Problem

Solution

See also

@@ Line 1: / Line 1: @@
 == 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.
-[[Image:AIME_83_-4.JPG]]
+<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 ==
 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>.
-[[Image:AIME_83_-4_Modified.JPG]]
+<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>.
-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 ==