1983 AIME Problems/Problem 4

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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.

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=in...

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 $x^2+y^2$ .

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...

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?

size(200);pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc;};pair a[] = {(0,0),(0,5...

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 $OB^2 = 26$ .

1983 AIME (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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