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

Revision as of 13:12, 31 January 2022

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.

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$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, $\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2$ , and $\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , such that the answer is $1^2 + 5^2 = \boxed{026}$ .

Solution 2

Drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ). Also, mark 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, so $AC = MO$ . Then, notice that $\angle MOT_2 = \angle T_3MO = \angle BAC$ . Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$ . As $T_3B = 3$ and $MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem tells us that $OB^2 = \boxed{026}$ .

Solution 3

Draw segment $OB$ with length $x$ , and draw 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 $\angle BAC=\arctan{\left(\frac{2}{6}\right)}$ and $\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}$ . 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 subtraction formula for $\tan$ , 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 $x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}$ $x^2=\boxed{026}.$

Solution 4

We use coordinates. Let the circle have center $(0,0)$ and radius $\sqrt{50}$ ; this circle has equation $x^2 + y^2 = 50$ . Let the coordinates of $B$ be $(a,b)$ . We want to find $a^2 + b^2$ . $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$ , respectively, both lie on the circle. From this we obtain the system of equations

$a^2 + (b+6)^2 = 50$

$(a+2)^2 + b^2 = 50$

After expanding these terms, we notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$ . after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$ . Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta.

Solving, we get $a=5$ and $b=-1$ , so the distance is $a^2 + b^2 = \boxed{026}$ .

@@ Line 88: / Line 88: @@
 <math>(a+2)^2 + b^2 = 50</math>
+After expanding these terms, we notice by subtracting the first and second equations, we can cancel out <math>a^2</math> and <math>b^2</math>. after substituting <math>a=3b+8</math> and plugging back in, we realize that <math>(a,b)=(-7,-5)</math> or <math>(5,-1)</math>. Since the first point is out of the circle, we find that <math>(5,-1)</math> is the only relevant answer. This paragraph is written by ~hastapasta.
 Solving, we get <math>a=5</math> and <math>b=-1</math>, so the distance is <math>a^2 + b^2 = \boxed{026}</math>.
 == See Also ==
 {{AIME box|year=1983|num-b=3|num-a=5}}
 [[Category:Intermediate Geometry Problems]]

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