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

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</asy>
 
</asy>
  
−
==Solution==
+
==Solution 1==
−
===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 <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.
 
Because we are given a right angle, we look for ways to apply the [[Pythagorean Theorem]].  Let the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math> be <math>D</math> and let the foot of the perpendicular from <math>O</math> to the [[line]] <math>BC</math> be <math>E</math>. Let <math>OE=x</math> and <math>OD=y</math>. We're trying to find <math>x^2+y^2</math>.
  
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Thus, <math>\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2</math>, and <math>\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.
 
Thus, <math>\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2</math>, and <math>\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2</math>. We solve this system to get <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.
  
−
===Solution 2===
+
==Solution 2==
 
We'll use the [[law of cosines]].  Let <math>O</math> be the center of the circle; we wish to find <math>OB</math>.  We know how long <math>OA</math> and <math>AB</math> are, so if we can find <math>\cos \angle OAB</math>, we'll be in good shape.
 
We'll use the [[law of cosines]].  Let <math>O</math> be the center of the circle; we wish to find <math>OB</math>.  We know how long <math>OA</math> and <math>AB</math> are, so if we can find <math>\cos \angle OAB</math>, we'll be in good shape.
  
Line 114: Line 113:
 
&= 26. \end{align*} </cmath>
 
&= 26. \end{align*} </cmath>
  
−
===Solution 3===
+
==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>).
 
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>.
 
Also, mark the midpoint <math>M</math> of <math>AC</math>.
Line 144: Line 143:
 
Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Therefore, the two blue triangles are congruent, from which we deduce <math>MT_2 = 2</math> and <math>OT_2 = 6</math>. As <math>T_3B = 3</math> and <math>MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem tells us that <math>OB^2 = \boxed{026}</math>.
 
Then, notice that <math>\angle MOT_2 = \angle T_3MO = \angle BAC</math>. Therefore, the two blue triangles are congruent, from which we deduce <math>MT_2 = 2</math> and <math>OT_2 = 6</math>. As <math>T_3B = 3</math> and <math>MT_3 = 1</math>, we subtract and get <math>OT_1 = 5,T_1B = 1</math>. Then the Pythagorean Theorem tells us that <math>OB^2 = \boxed{026}</math>.
  
−
===Solution 4===
+
==Solution 4==
 
Draw segment <math>OB</math> with length <math>x</math>, and draw radius <math>OQ</math> such that <math>OQ</math> bisects chord <math>AC</math> at point <math>M</math>. This also means that <math>OQ</math> is perpendicular to <math>AC</math>. By the Pythagorean Theorem, we get that <math>AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}</math>, and therefore <math>AM=\sqrt{10}</math>. Also by the Pythagorean theorem, we can find that <math>OM=\sqrt{50-10}=2\sqrt{10}</math>.
 
Draw segment <math>OB</math> with length <math>x</math>, and draw radius <math>OQ</math> such that <math>OQ</math> bisects chord <math>AC</math> at point <math>M</math>. This also means that <math>OQ</math> is perpendicular to <math>AC</math>. By the Pythagorean Theorem, we get that <math>AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}</math>, and therefore <math>AM=\sqrt{10}</math>. Also by the Pythagorean theorem, we can find that <math>OM=\sqrt{50-10}=2\sqrt{10}</math>.
  
 
Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath>
 
Next, find <math>\angle BAC=\arctan{\left(\frac{2}{6}\right)}</math> and <math>\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}</math>. Since <math>\angle OAB=\angle OAM-\angle BAC</math>, we get <cmath>\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}</cmath>By the subtraction formula for <math>\tan</math>, we get<cmath>\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}</cmath><cmath>\tan{(\angle OAB)}=1</cmath><cmath>\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}</cmath>Finally, by the Law of Cosines on <math>\triangle OAB</math>, we get <cmath>x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}</cmath><cmath>x^2=\boxed{026}.</cmath>
  
−
===Solution 5===
+
==Solution 5==
 
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
  
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Solving, we get <math>a=5</math> and <math>b=-1</math>, so the distance is <math>a^2 + b^2 = \boxed{026}</math>.
 
Solving, we get <math>a=5</math> and <math>b=-1</math>, so the distance is <math>a^2 + b^2 = \boxed{026}</math>.
 +
 +
== Solution 6 ==
 +
<center><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);
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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></center>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See Also ==
 
== See Also ==

Revision as of 22:55, 16 July 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); 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

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

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 $AC$.

[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

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

[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]

~isabelchen

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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