Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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In <math>\triangle ABC</math>, <math>AB = 86</math>, and <math>AC=97</math>. A circle with center <math>A</math> and radius <math>AB</math> intersects <math>\overline{BC}</math> at points <math>B</math> and <math>X</math>. Moreover <math>\overline{BX}</math> and <math>\overline{CX}</math> have integer lengths. What is <math>BC</math>? | In <math>\triangle ABC</math>, <math>AB = 86</math>, and <math>AC=97</math>. A circle with center <math>A</math> and radius <math>AB</math> intersects <math>\overline{BC}</math> at points <math>B</math> and <math>X</math>. Moreover <math>\overline{BX}</math> and <math>\overline{CX}</math> have integer lengths. What is <math>BC</math>? | ||
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<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 </math> | <math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 </math> | ||
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− | + | ==Solution 1 (Number Theoretic Power of a Point)== | |
− | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meets 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 (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, so <math>p</math> is either 3,11, or 33. We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>. <math>p</math> is 33 | + | Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meets 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 (Secant-Secant Power Theorem), we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>. We know that <math>p+q>p</math>, so <math>p</math> is either <math>3</math>, <math>11</math>, or <math>33</math>. We also know that <math>p>11</math> by the triangle inequality on <math>\triangle ACX</math>. Thus, <math>p</math> is <math>33</math> so we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>. |
− | + | ==Solution 2 ([[Stewart's Theorem]])== | |
− | [[Stewart's Theorem]] | ||
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>. | 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>. | ||
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<math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math> | <math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math> | ||
+ | |||
+ | <math>x^2 y + xy^2 + 86^2 y = 97^2 y</math> | ||
<math>x^2 + xy + 86^2 = 97^2</math> | <math>x^2 + xy + 86^2 = 97^2</math> | ||
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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>. | 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 3== | |
Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>. | Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>. | ||
Use power of a point on point C to the circle centered at A. | Use power of a point on point C to the circle centered at A. | ||
− | So <math>CX | + | So <math>CX \cdot CB=CD \cdot CE \Rightarrow x(x+y)=(97-86)(97+86) \Rightarrow x(x+y)=3*11*61</math>. |
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Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>. | Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>. | ||
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Therefore, the answer is <math> \boxed{\textbf{(D) }61}</math>. | Therefore, the answer is <math> \boxed{\textbf{(D) }61}</math>. | ||
− | + | ==Solution 4== | |
− | + | <asy> | |
unitsize(2); | unitsize(2); | ||
import olympiad; | import olympiad; | ||
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fill(anglemark(A,E,D,100),black); | fill(anglemark(A,E,D,100),black); | ||
− | label(" | + | label("$90^\circ$",anglemark(A,E,D),3*S); |
− | + | </asy> | |
− | We first draw the height of isosceles triangle ABD | + | We first draw the height of isosceles triangle <math>ABD</math> and get two equations by the [[Pythagorean Theorem]]. |
First, <math>h^2 + k^2 = 86^2</math>. Second, <math>h^2 + (k + m)^2 = 97^2</math>. | First, <math>h^2 + k^2 = 86^2</math>. Second, <math>h^2 + (k + m)^2 = 97^2</math>. | ||
Subtracting these two equations, we get <math>2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013</math>. | Subtracting these two equations, we get <math>2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013</math>. | ||
We then add <math>k^2</math> to both sides to get <math>k^2 + 2km + m^2 = 2013 + k^2</math>. | We then add <math>k^2</math> to both sides to get <math>k^2 + 2km + m^2 = 2013 + k^2</math>. | ||
We then complete the square to get <math>(k + m)^2 = 2013 + k^2</math>. Because <math>k</math> and <math>m</math> are both integers, we get that <math>2013 + k^2</math> is a square number. Simple guess and check reveals that <math>k = 14</math>. | We then complete the square to get <math>(k + m)^2 = 2013 + k^2</math>. Because <math>k</math> and <math>m</math> are both integers, we get that <math>2013 + k^2</math> is a square number. Simple guess and check reveals that <math>k = 14</math>. | ||
− | Because <math>k</math> equals <math>14</math>, therefore <math>m = 33</math>. We want <math>\overline{BC} = 2k + m</math>, so we get that <math>\overline{BC} = \boxed{( | + | Because <math>k</math> equals <math>14</math>, therefore <math>m = 33</math>. We want <math>\overline{BC} = 2k + m</math>, so we get that <math>\overline{BC} = \boxed{(D)~61}</math>. |
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+ | <math>\phantom{solution and diagram by bobjoe123}</math> | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>BX.</math> Since <math>\triangle ABX</math> is isosceles <math>AX=AB=86,EB=EX,</math> and the answer is <math>EC+EB=EC+EX.</math> <math>(EC+EX)(EC-EX)=EC^2-EX^2=(97^2-AE^2)-(86^2-AE^2)=97^2-86^2=2013</math> by the Pythagorean Theorem. Only <math>EC+EX=\boxed{(D)~61}</math> is a factor of <math>2013</math> such that <math>97>EC+EX>EC-EX=\frac{2013}{EC+EX}.</math> | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://www.youtube.com/watch?v=f1nxu8MWWKc | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=2692 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Number theory]] | ||
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:02, 21 August 2023
- The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
In ,
, and
. A circle with center
and radius
intersects
at points
and
. Moreover
and
have integer lengths. What is
?
Solution 1 (Number Theoretic Power of a Point)
Let ,
, and
meets the circle at
and
, with
on
. Then
. Using the Power of a Point (Secant-Secant Power Theorem), we get that
. We know that
, so
is either
,
, or
. We also know that
by the triangle inequality on
. Thus,
is
so we get that
.
Solution 2 (Stewart's Theorem)
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 3
Let . Let the circle intersect
at
and the diameter including
intersect the circle again at
.
Use power of a point on point C to the circle centered at A.
So .
Obviously so we have three solution pairs for
.
By the Triangle Inequality, only
yields a possible length of
.
Therefore, the answer is .
Solution 4
We first draw the height of isosceles triangle and get two equations by the Pythagorean Theorem.
First,
. Second,
.
Subtracting these two equations, we get
.
We then add
to both sides to get
.
We then complete the square to get
. Because
and
are both integers, we get that
is a square number. Simple guess and check reveals that
.
Because
equals
, therefore
. We want
, so we get that
.
Solution 5
Let be the foot of the altitude from
to
Since
is isosceles
and the answer is
by the Pythagorean Theorem. Only
is a factor of
such that
~dolphin7
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=f1nxu8MWWKc
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=2692
~ pi_is_3.14
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.