1983 AIME Problems/Problem 4

Revision as of 21:37, 12 June 2018 by Soyamyboya (talk | contribs) (Solution)

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<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 \[x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}\]\[x^2=\boxed{026}\]

See Also

1983 AIME (ProblemsAnswer KeyResources)
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