1983 AIME Problems/Problem 4: Difference between revisions

Latest revision as of 19:54, 31 December 2024

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.

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]$

Solution 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 $x^2+y^2$ .

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label("$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 (Trig)

We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$ . We know how long $OA$ and $AB$ are, so if we can find $\cos \angle OAB$ , we'll be in good shape.

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]$

We can find $\cos \angle OAB$ using angles $OAC$ and $BAC$ . First we note that by Pythagoras, $AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}.$ If we let $M$ be the midpoint of $AC$ , that mean that $AM = \sqrt{10}$ . Since $\triangle OAC$ is isosceles ( $OA = OC$ from the definition of a circle), $M$ is also the foot of the altitude from $O$ to $AC.$

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$M$",M,SSW); label("$C$",C,SE); label("$\sqrt{50}$", (O+A)/2, NW); label("$\sqrt{10}$", (A+M)/2, E); [/asy]$

It follows that $OM = \sqrt{40} = 2 \sqrt{10}$ . Therefore $\begin{align*} \cos \angle OAC = \frac{\sqrt{10}}{\sqrt{50}} &= \frac{1}{\sqrt{5}}, \\ \sin \angle OAC = \frac{2 \sqrt{10}}{\sqrt{50}} &= \frac{2}{\sqrt{5}}. \end{align*}$ Meanwhile, from right triangle $ABC,$ we have $\begin{align*} \cos \angle BAC = \frac{6}{\sqrt{40}} &= \frac{3}{\sqrt{10}}, \\ \sin \angle BAC = \frac{2}{\sqrt{40}} &= \frac{1}{\sqrt{10}}. \end{align*}$

This means that by the angle subtraction formulas, $\begin{align*} \cos \angle OAB &= \cos (\angle OAC - \angle BAC) \\ &= \cos \angle OAC \cos \angle BAC + \sin \angle OAC \sin \angle BAC \\ &= \frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} \\ &= \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}}. \end{align*}$

Now we have all we need to use the law of cosines on $\triangle OAB.$ This tells us that $\begin{align*} OB^2 &= AO^2 + AB^2 - 2 AO \cdot AB \cdot \cos \angle OAB \\ &= 50 + 36 - 2 \cdot 5 \sqrt{2} \cdot 6 \cdot \frac{1}{\sqrt{2}} \\ &= 86 - 2 \cdot 5 \cdot 6 \\ &= 26. \end{align*}$

Solution 3

Mark the midpoint $M$ of $AC$ . Then, drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ).

$[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {"O","$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 4 (Trig)

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 5

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 $a^2 + b^2 = \boxed{026}$ .

Solution 6 (Trigonometry)

$[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); draw(O--C); dot(O); dot(A); dot(B); dot(C); label("$O$",O,SW); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]$

I will use the law of cosines in triangle $\triangle OAC$ and $\triangle OBC$ .

$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}$

$\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}$

$\cos \angle ACO = \frac{AC^2+OC^2-OA^2}{2 \cdot AC \cdot OC} = \frac{(2\sqrt{10})^2+(\sqrt{50})^2-(\sqrt{50})^2}{2 \cdot 2\sqrt{10} \cdot \sqrt{50}} = \frac{1}{\sqrt{5}}$

$\sin \angle ACB = \sqrt{1-\cos^2 \angle ACB} = \sqrt{1-(\frac{1}{\sqrt{10}})^2} = \frac{3}{\sqrt{10}}$

$\sin \angle ACO = \sqrt{1-\cos^2 \angle ACO} = \sqrt{1-(\frac{1}{\sqrt{5}})^2} = \frac{2}{\sqrt{5}}$

$\cos \angle OCB = \cos (\angle ACB - \angle ACO) = \cos \angle ACB \cdot \cos \angle ACO + \sin \angle ACB \cdot \sin \angle ACO = \frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{3}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{7}{5\sqrt{2}}$

$OB^2 = OC^2 + BC^2 - 2 \cdot OC \cdot BC \cdot \cos \angle OCB = (\sqrt{50})^2 + 2^2 - 2 \cdot \sqrt{50} \cdot 2 \cdot \frac{7}{5\sqrt{2}} = 50 + 4 - 28 = \boxed{026}$

~isabelchen

Solution 7

Notice that $50=5^2+5^2=7^2+1^2$ , and by the size of the diagram, it seems reasonable that $OA$ represents $5^2+5^2$ , and $OC$ means the $7^1+1^2$ , and indeed, the values work ( $7-5=2$ and $5+1=6$ ), so $OB^2=5^2+1^2=\boxed{026}$

Note: THIS IS NOT A RELIABLE SOLUTION, as diagrams on the tests are not usually drawn to scale.

Video Solution by Pi Academy

https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D

~ Pi Academy

@@ Line 114: / Line 114: @@
 ==Solution 3==
-Drop perpendiculars from <math>O</math> to <math>AB</math> (with foot <math>T_1</math>), <math>M</math> to <math>OT_1</math> (with foot <math>T_2</math>), and <math>M</math> to <math>AB</math> (with foot <math>T_3</math>).
+Mark the midpoint <math>M</math> of <math>AC</math>. Then, drop perpendiculars from <math>O</math> to <math>AB</math> (with foot <math>T_1</math>), <math>M</math> to <math>OT_1</math> (with foot <math>T_2</math>), and <math>M</math> to <math>AB</math> (with foot <math>T_3</math>).
-Also, mark the midpoint <math>M</math> of <math>AC</math>.
 <center><asy>

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