Difference between revisions of "1983 AIME Problems/Problem 4"
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− | == Problem == | + | ==Problem== |
− | A machine shop cutting tool | + | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math>B</math> 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 <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>. | ||
− | + | <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); | ||
+ | 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></center><!-- Asymptote replacement for Image:AIME_83_-4_Modified.JPG by bpms --> | ||
− | + | Applying the Pythagorean Theorem, <math>OA^2 = OD^2 + AD^2</math> and <math>OC^2 = EC^2 + EO^2</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 (Trig)== |
− | * [[ | + | 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. |
− | + | ||
− | * [[ | + | <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); | ||
+ | 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></center> | ||
+ | |||
+ | We can find <math>\cos \angle OAB</math> using angles <math>OAC</math> and <math>BAC</math>. First we note that by [[Pythagorean theorem | Pythagoras]], | ||
+ | <cmath> AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}. </cmath> | ||
+ | If we let <math>M</math> be the midpoint of <math>AC</math>, that mean that <math>AM = \sqrt{10}</math>. Since <math>\triangle OAC</math> is isosceles (<math>OA = OC</math> from the definition of a circle), <math>M</math> is also the foot of the altitude from <math>O</math> to <math>AC.</math> | ||
+ | |||
+ | <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); | ||
+ | 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></center> | ||
+ | It follows that <math>OM = \sqrt{40} = 2 \sqrt{10}</math>. Therefore | ||
+ | <cmath> \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*}</cmath> | ||
+ | Meanwhile, from right triangle <math>ABC,</math> we have | ||
+ | <cmath> \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*} </cmath> | ||
+ | |||
+ | This means that by the [[Trigonometric_identities#Angle_addition_identities | angle subtraction formulas]], | ||
+ | <cmath> \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*} </cmath> | ||
+ | |||
+ | Now we have all we need to use the law of cosines on <math>\triangle OAB.</math> This tells us that | ||
+ | <cmath> \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*} </cmath> | ||
+ | |||
+ | ==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>). | ||
+ | Also, mark the midpoint <math>M</math> of <math>AC</math>. | ||
+ | |||
+ | <center><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></center> | ||
+ | First notice that by computation, <math>OAC</math> is a <math>\sqrt {50} - \sqrt {40} - \sqrt {50}</math> isosceles triangle, so <math>AC = MO</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 (Trig)== | ||
+ | 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> | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | <math>a^2 + (b+6)^2 = 50</math> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | == Solution 6 (Trigonometry)== | ||
+ | <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); | ||
+ | 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> | ||
+ | |||
+ | I will use the [[law of cosines]] in triangle <math>\triangle OAC</math> and <math>\triangle OBC</math>. | ||
+ | |||
+ | <math>AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}</math> | ||
+ | |||
+ | <math>\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}</math> | ||
+ | |||
+ | <math>\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}}</math> | ||
+ | |||
+ | <math>\sin \angle ACB = \sqrt{1-\cos^2 \angle ACB} = \sqrt{1-(\frac{1}{\sqrt{10}})^2} = \frac{3}{\sqrt{10}}</math> | ||
+ | |||
+ | <math>\sin \angle ACO = \sqrt{1-\cos^2 \angle ACO} = \sqrt{1-(\frac{1}{\sqrt{5}})^2} = \frac{2}{\sqrt{5}}</math> | ||
+ | |||
+ | <math>\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}}</math> | ||
+ | |||
+ | <math>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}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 7== | ||
+ | Notice that <math>50=5^2+5^2=7^2+1^2</math>, and by the size of the diagram, it seems reasonable that <math>OA</math> represents <math>5^2+5^2</math>, and <math>OC</math> means the <math>7^1+1^2</math>, and indeed, the values work (<math>7-5=2</math> and <math>5+1=6</math>), so <math>OB^2=5^2+1^2=\boxed{026}</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == See Also == | ||
+ | {{AIME box|year=1983|num-b=3|num-a=5}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 20:07, 8 December 2024
Contents
[hide]Problem
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is cm, the length of is cm and that of is cm. The angle is a right angle. Find the square of the distance (in centimeters) from to the center of the circle.
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 to be and let the foot of the perpendicular from to the line be . Let and . We're trying to find .
Applying the Pythagorean Theorem, and .
Thus, , and . We solve this system to get and , such that the answer is .
Solution 2 (Trig)
We'll use the law of cosines. Let be the center of the circle; we wish to find . We know how long and are, so if we can find , we'll be in good shape.
We can find using angles and . First we note that by Pythagoras, If we let be the midpoint of , that mean that . Since is isosceles ( from the definition of a circle), is also the foot of the altitude from to
It follows that . Therefore Meanwhile, from right triangle we have
This means that by the angle subtraction formulas,
Now we have all we need to use the law of cosines on This tells us that
Solution 3
Drop perpendiculars from to (with foot ), to (with foot ), and to (with foot ). Also, mark the midpoint of .
First notice that by computation, is a isosceles triangle, so . Then, notice that . Therefore, the two blue triangles are congruent, from which we deduce and . As and , we subtract and get . Then the Pythagorean Theorem tells us that .
Solution 4 (Trig)
Draw segment with length , and draw radius such that bisects chord at point . This also means that is perpendicular to . By the Pythagorean Theorem, we get that , and therefore . Also by the Pythagorean theorem, we can find that .
Next, find and . Since , we get By the subtraction formula for , we getFinally, by the Law of Cosines on , we get
Solution 5
We use coordinates. Let the circle have center and radius ; this circle has equation . Let the coordinates of be . We want to find . and with coordinates and , respectively, both lie on the circle. From this we obtain the system of equations
After expanding these terms, we notice by subtracting the first and second equations, we can cancel out and . after substituting and plugging back in, we realize that or . Since the first point is out of the circle, we find that is the only relevant answer. This paragraph is written by ~hastapasta.
Solving, we get and , so the distance is .
Solution 6 (Trigonometry)
I will use the law of cosines in triangle and .
Solution 7
Notice that , and by the size of the diagram, it seems reasonable that represents , and means the , and indeed, the values work ( and ), so
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
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 |