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

Revision as of 10:58, 30 September 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$",C,SE); [/asy]$

Solutions

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 $O$ 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 $1^2 + 5^2 = \boxed{026}$ .

Solution 2: Synthetic Solution

Drop perpendiculars from $O$ to $AB$ ( $T_1$ ), $M$ to $OT_1$ ( $T_2$ ), and $M$ to $AB$ ( $T_3$ ). Also, draw the midpoint $M$ of $AC$ .

Then the problem is trivialized. Why?

$[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"}; for(int i=0;i<a.length;++i) { dl(n[i],a[i],dir(degrees(a[i],false) ) ); draw(a[(i-1)%a.length]--a[i]); }; dot(a); draw(a[5]--a[1]); draw(a[0]--a[3]); draw(a[0]--a[4]); draw(a[0]--a[2]); draw(a[0]--a[5]); draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7)); draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7)); [/asy]$

First notice that by computation, $OAC$ is a $\sqrt {50} - \sqrt {40} - \sqrt {50}$ isosceles triangle; thus $AC = MO$ . Then, notice that $\angle MOT_2 = \angle T_3MO = \angle BAC$ . Thus the two blue triangles are congruent.

So, $MT_2 = 2,OT_2 = 6$ . As $T_3B = 3, MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem shows $OB^2 = 26$ .

@@ Line 15: / Line 15: @@
 label("$C$",C,SE);
 </asy></center>
+__TOC__
 == Solutions ==
@@ Line 47: / Line 49: @@
 Then the problem is trivialized. Why?
-<asy>
+<center><asy>
-unitsize(1.5cm);
+size(200);
 pair dl(string name, pair loc, pair offset) {
   dot(loc);
@@ Line 67: / Line 69: @@
 draw(a[0]--a[5]);
-draw(a[5]--a[2]--a[3]--cycle,blue+.7mm);
+draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7));
-draw(a[0]--a[8]--a[7]--cycle,blue+.7mm);
+draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7));
-</asy>
+</asy></center>
 First notice that by computation, <math>OAC</math> is a <math>\sqrt {50} - \sqrt {40} - \sqrt {50}</math> isosceles triangle; thus <math>AC = MO</math>.
 Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Thus the two blue triangles are congruent.

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

Page

Toolbox

Search