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Difference between revisions of "1983 AIME Problems/Problem 4"
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===Solution 4===
 
===Solution 4===
 
We use coordinates. Let the circle have center <math>(0,0)</math> and radius <math>\sqrt{50}</math>; this circle has equation <math>x^2 + y^2 = 50</math>. Let the coordinates of <math>B</math> be <math>(a,b)</math>. We want to find <math>a^2 + b^2</math>. <math>A</math> and <math>C</math> with coordinates <math>(a,b+6)</math> and <math>(a+2,b)</math>, respectively, both lie on the circle. From this we obtain the system of equations
 
We use coordinates. Let the circle have center <math>(0,0)</math> and radius <math>\sqrt{50}</math>; this circle has equation <math>x^2 + y^2 = 50</math>. Let the coordinates of <math>B</math> be <math>(a,b)</math>. We want to find <math>a^2 + b^2</math>. <math>A</math> and <math>C</math> with coordinates <math>(a,b+6)</math> and <math>(a+2,b)</math>, respectively, both lie on the circle. From this we obtain the system of equations
 +
 
<math>a^2 + (b+6)^2 = 50</math>
 
<math>a^2 + (b+6)^2 = 50</math>
  

Revision as of 21:42, 30 October 2020

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] Pdfresizer.com-pdf-convert-aimeq4.png

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

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

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

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

See Also

1983 AIME (Problems • Answer Key • Resources)
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
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