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

Revision as of 18:11, 15 February 2019

Problem

A machine-shop cutting tool has 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.

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$"}; for(int i=0;i<a.length;++i) { dl(n[i],a[i],dir(degrees(a[i],false) ) ); draw(a[(i-1)%a.length]--a[i]); }; dot(a); draw(a[5]--a[1]); draw(a[0]--a[3]); draw(a[0]--a[4]); draw(a[0]--a[2]); draw(a[0]--a[5]); draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7)); draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7)); [/asy]$

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

Solution 3

Draw segment $OB$ with length $x$ and radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$ . This also means that $OQ$ is perpendicular to $AC$ . By the Pythagorean Theorem, we get that $AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}$ , and therefore $AM=\sqrt{10}$ . Also by the Pythagorean theorem, we can find that $OM=\sqrt{50-10}=2\sqrt{10}$ .

Next, find $\angle BAC=\arctan{(\frac{2}{6})}$ and $\angle OAM=\arctan{(\frac{2\sqrt{10}}{\sqrt{10}})}$ . Since $\angle OAB=\angle OAM-\angle BAC$ , we get $\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}$ $\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}$ By the tangent angle subtraction formula, we get $\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}$ $\tan{(\angle OAB)}=1$ $\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}$ Finally, by the Law of Cosines on $\triangle OAB$ , we get $x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}$ $x^2=\boxed{026}$

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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