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

## 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

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

### Solution 2

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

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 $OB^2 = \boxed{026}$.