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

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Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = \boxed{026}</math>.
 
Thus, <math>(\sqrt{50})^2 = y^2 + (6-x)^2</math>, and <math>(\sqrt{50})^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, resulting in an answer of <math>1^2 + 5^2 = \boxed{026}</math>.
 +
== Synthetic Solution ==
 +
Drop perpendiculars from <math>O</math> to <math>AB</math> (<math>T_1</math>), <math>M</math> to <math>OT_1</math> (<math>T_2</math>), and <math>M</math> to <math>AB</math> (<math>T_3</math>).
 +
Also, draw the midpoint <math>M</math> of <math>AC</math>.
 +
 +
Then the problem is trivialized. Why?
 +
<asy>
 +
unitsize(1.5cm);
 +
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+.7mm);
 +
draw(a[0]--a[8]--a[7]--cycle,blue+.7mm);
 +
</asy>
 +
First notice that by computation, <math>OAC</math> is a <math>\sqrt {50} - \sqrt {40} - \sqrt {50}</math> isosceles triangle; thus <math>AC = MO</math>.
 +
Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Thus the two blue triangles are congruent.
 +
 +
So, <math>MT_2 = 2,OT_2 = 6</math>. As <math>T_3B = 3, MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem shows <math>OB^2 = 26</math>.
  
 
== See also ==
 
== See also ==

Revision as of 03:21, 30 September 2008

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

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 an answer of $1^2 + 5^2 = \boxed{026}$.

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? [asy] unitsize(1.5cm); 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+.7mm); draw(a[0]--a[8]--a[7]--cycle,blue+.7mm); [/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 $OB^2 = 26$.

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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