Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | ===Solution 1=== | + | ===Solution 1 (Diophantine PoP)=== |
− | + | <asy> | |
− | + | //Made by samrocksnature | |
+ | size(8cm); | ||
+ | pair A,B,C,D,E,X; | ||
+ | A=(0,0); | ||
+ | B=(-53.4,-67.4); | ||
+ | C=(0,-97); | ||
+ | D=(0,-86); | ||
+ | E=(0,86); | ||
+ | X=(-29,-81); | ||
+ | draw(circle(A,86)); | ||
+ | draw(E--C--B--A--X); | ||
+ | label("$A$",A,NE); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,NE); | ||
+ | label("$X$",X,dir(250)); | ||
+ | dot(A^^B^^C^^D^^E^^X); | ||
+ | </asy> | ||
− | + | Let circle <math>A</math> intersect <math>AC</math> at <math>D</math> and <math>E</math> as shown. We apply Power of a Point on point <math>C</math> with respect to circle <math>A.</math> This yields the diophantine equation | |
− | |||
− | |||
− | + | <cmath>CX \cdot CB = CD \cdot CE</cmath> | |
− | + | <cmath>CX(CX+XB) = (97-86)(97+86)</cmath> | |
+ | <cmath>CX(CX+XB) = 3 \cdot 11 \cdot 61.</cmath> | ||
− | + | Since lengths cannot be negative, we must have <math>CX+XB \ge CX.</math> This generates the four solution pairs for <math>(CX,CX+XB)</math>: <cmath>(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).</cmath> | |
+ | However, by the Triangle Inequality on <math>\triangle ACX,</math> we see that <math>CX>13.</math> This implies that we must have <math>CX+XB= \boxed{\textbf{(D) }61}.</math> | ||
+ | |||
+ | (Solution by unknown, latex/asy modified majorly by samrocksnature) | ||
===Solution 2=== | ===Solution 2=== | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | + | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>AC</math>. Then <math>AZ = AY = 86</math>. Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math> | |
− | |||
− | |||
===Solution 3=== | ===Solution 3=== | ||
− | Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB | + | Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB = AX = 86</math>. |
Then by Stewart's Theorem, | Then by Stewart's Theorem, | ||
Line 43: | Line 56: | ||
<math>x^2 + xy + 86^2 = 97^2</math> | <math>x^2 + xy + 86^2 = 97^2</math> | ||
− | (Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.) | + | (Since <math>y</math> cannot be equal to <math>0</math>, dividing both sides of the equation by <math>y</math> is allowed.) |
<math>x(x+y) = (97+86)(97-86)</math> | <math>x(x+y) = (97+86)(97-86)</math> | ||
Line 49: | Line 62: | ||
<math>x(x+y) = 2013</math> | <math>x(x+y) = 2013</math> | ||
− | The prime factors of 2013 are 3, 11, and 61. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal 33, and <math>x+y</math> must equal | + | The prime factors of <math>2013</math> are <math>3</math>, <math>11</math>, and <math>61</math>. Obviously, <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal <math>33</math>, and <math>x+y</math> must equal <math> \boxed{\textbf{(D) }61}</math> |
+ | ===Solution 4=== | ||
+ | Motivation and general line of reasoning: we use a law of cosines condition on triangle <math>ABX</math> and triangle <math>ABC</math> to derive some equivalent formations of the same quantity <math>\cos B</math>, which looks promising since it involves the desired length <math>BC</math>, as well as <math>BX</math> and <math>CX</math>.An intermediate step would be to use the integer condition and pay attention to the divisors of stuff. | ||
+ | |||
+ | Let <math>BX</math>=<math>x_1</math> and <math>CX</math>=<math>x_2</math>. | ||
+ | |||
+ | First we have | ||
+ | <math>\cos(B)=\frac{86^2+x_{1}^2-86^2}{172x_{1}}</math> | ||
+ | by applying the law of cosines to triangle <math>ABX</math>. Another equivalent formation of <math>\cos B</math> is | ||
+ | <math>\frac{86^2+(x_{1}+x_2)^2-97^2}{172(x_{1}+x_2)}</math>. | ||
+ | Now that we have the necessary ingredients, we can make a system of equations and deduce that | ||
+ | <math>x_1=\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}</math>. | ||
+ | |||
+ | Don't lose focus by now-we try to find <math>x_1+x_2</math>, an integer value as given in the problem. To do this, we want the quantity | ||
+ | <math>\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}</math> | ||
+ | to 1) be an integer and 2) smaller than <math>x_1+x_2</math>. For the sake of conciseness in notation we let <math>M=x_1+x_2</math>, then <math>M+\frac{86^2-97^2}{M}</math> is an integer. Now recalling the fact that <math>(a+b)(a-b)=a^2-b^2</math>, we get that <math>\frac{183(-11)}{M}</math> must be an integer. | ||
+ | |||
+ | Now the prime factor decomposition of <math>183 \cdot -11</math> is <math>(61)(3)(-11)</math>. Trying out all the possible integer values that divide this quantity, we get that the only viable option for <math>M</math> is 61 (verify that yourself!) Therefore the answer is <math> \boxed{\textbf{(D) }61}</math>. (Solution by CreamyCream123) | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc12a/357 | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/zxW3uvCQFls | ||
+ | |||
+ | ~sugar_rush | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Number theory]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:16, 3 December 2024
Contents
[hide]Problem
In , , and . A circle with center and radius intersects at points and . Moreover and have integer lengths. What is ?
Solution
Solution 1 (Diophantine PoP)
Let circle intersect at and as shown. We apply Power of a Point on point with respect to circle This yields the diophantine equation
Since lengths cannot be negative, we must have This generates the four solution pairs for :
However, by the Triangle Inequality on we see that This implies that we must have
(Solution by unknown, latex/asy modified majorly by samrocksnature)
Solution 2
Let , , and meet the circle at and , with on . Then . Using the Power of a Point, we get that . We know that , and that by the triangle inequality on . Thus, we get that
Solution 3
Let represent , and let represent . Since the circle goes through and , . Then by Stewart's Theorem,
(Since cannot be equal to , dividing both sides of the equation by is allowed.)
The prime factors of are , , and . Obviously, . In addition, by the Triangle Inequality, , so . Therefore, must equal , and must equal
Solution 4
Motivation and general line of reasoning: we use a law of cosines condition on triangle and triangle to derive some equivalent formations of the same quantity , which looks promising since it involves the desired length , as well as and .An intermediate step would be to use the integer condition and pay attention to the divisors of stuff.
Let = and =.
First we have by applying the law of cosines to triangle . Another equivalent formation of is . Now that we have the necessary ingredients, we can make a system of equations and deduce that .
Don't lose focus by now-we try to find , an integer value as given in the problem. To do this, we want the quantity to 1) be an integer and 2) smaller than . For the sake of conciseness in notation we let , then is an integer. Now recalling the fact that , we get that must be an integer.
Now the prime factor decomposition of is . Trying out all the possible integer values that divide this quantity, we get that the only viable option for is 61 (verify that yourself!) Therefore the answer is . (Solution by CreamyCream123)
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/357
~dolphin7
Video Solution
~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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