Difference between revisions of "2014 AMC 12B Problems/Problem 24"
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\textbf{(E) }421\qquad</math> | \textbf{(E) }421\qquad</math> | ||
− | ==Solution== | + | ==Video Solution by Punxsutawney Phil== |
+ | https://www.youtube.com/watch?v=1-2vT_GIceA | ||
− | + | == Solution 1 == | |
− | + | Let <math>BE=a</math>, <math>AD=b</math>, and <math>AC=CE=BD=c</math>. Let <math>F</math> be on <math>AE</math> such that <math>CF \perp AE</math>. | |
+ | <asy> | ||
+ | size(200); | ||
+ | defaultpen(linewidth(0.4)+fontsize(10)); | ||
+ | pen s = linewidth(0.8)+fontsize(8); | ||
− | + | pair O,A,B,C,D,E0,F; | |
+ | O=origin; | ||
+ | A= dir(198); | ||
+ | path c = CR(O,1); | ||
+ | real r = 0.13535; | ||
+ | B = IP(c, CR(A,3*r)); | ||
+ | C = IP(c, CR(B,10*r)); | ||
+ | D = IP(c, CR(C,3*r)); | ||
+ | E0 = OP(c, CR(D,10*r)); | ||
+ | F = foot(C,A,E0); | ||
− | + | dot("$A$", A, A-O); | |
+ | dot("$B$", B, B-O); | ||
+ | dot("$C$", C, C-O); | ||
+ | dot("$D$", D, D-O); | ||
+ | dot("$E$", E0, E0-O); | ||
+ | dot("$F$", F, F-C); | ||
+ | label("$c$",A--C,S); | ||
+ | label("$c$",E0--C,W); | ||
+ | label("$7$",F--E0,S); | ||
+ | label("$7$",F--A,S); | ||
+ | label("$3$",A--B,2*W); | ||
+ | label("$10$",B--C,2*N); | ||
+ | label("$3$",C--D,2*NE); | ||
+ | label("$10$",D--E0,E); | ||
+ | draw(A--B--C--D--E0--A, black+0.8); | ||
− | <math> | + | draw(CR(O,1), s); |
+ | draw(A--C--E0, royalblue); | ||
+ | draw(C--F, royalblue+dashed); | ||
+ | draw(rightanglemark(E0,F,C,2)); | ||
+ | MA("\theta",A,B,C,0.075); | ||
+ | MA("\pi-\theta",C,E0,A,0.1); | ||
+ | </asy> | ||
+ | In <math>\triangle CFE</math> we have <math>\cos\theta = -\cos(\pi-\theta)=-7/c</math>. We use the [[Law of Cosines]] on <math>\triangle ABC</math> to get <math>60\cos\theta = 109-c^2</math>. Eliminating <math>\cos\theta</math> we get <math>c^3-109c-420=0</math> which factorizes as | ||
+ | <cmath>(c+7)(c+5)(c-12)=0.</cmath>Discarding the negative roots we have <math>c=12</math>. Thus <math>BD=AC=CE=12</math>. For <math>BE=a</math>, we use Ptolemy's theorem on cyclic quadrilateral <math>ABCE</math> to get <math>a=44/3</math>. For <math>AD=b</math>, we use [[Ptolemy's theorem]] on cyclic quadrilateral <math>ACDE</math> to get <math>b=27/2</math>. | ||
− | + | The sum of the lengths of the diagonals is <math>12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}</math> so the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math> | |
− | * | + | == Solution 2 == |
+ | Let <math>a</math> denote the length of a diagonal opposite adjacent sides of length <math>14</math> and <math>3</math>, <math>b</math> for sides <math>14</math> and <math>10</math>, and <math>c</math> for sides <math>3</math> and <math>10</math>. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | c^2 &= 3a+100 \\ | ||
+ | c^2 &= 10b+9 \\ | ||
+ | ab &= 30+14c \\ | ||
+ | ac &= 3c+140\\ | ||
+ | bc &= 10c+42 | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | |||
+ | Using equations <math>(1)</math> and <math>(2)</math>, we obtain: | ||
+ | |||
+ | <cmath> | ||
+ | a = \frac{c^2-100}{3} | ||
+ | </cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath> | ||
+ | b = \frac{c^2-9}{10} | ||
+ | </cmath> | ||
+ | |||
+ | Plugging into equation <math>(4)</math>, we find that: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{c^2-100}{3}c &= 3c + 140\\ | ||
+ | \frac{c^3-100c}{3} &= 3c + 140\\ | ||
+ | c^3-100c &= 9c + 420\\ | ||
+ | c^3-109c-420 &=0\\ | ||
+ | (c-12)(c+7)(c+5)&=0 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Or similarly into equation <math>(5)</math> to check: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{c^2-9}{10}c &= 10c+42\\ | ||
+ | \frac{c^3-9c}{10} &= 10c + 42\\ | ||
+ | c^3-9c &= 100c + 420\\ | ||
+ | c^3-109c-420 &=0\\ | ||
+ | (c-12)(c+7)(c+5)&=0 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <math>c</math>, being a length, must be positive, implying that <math>c=12</math>. In fact, this is reasonable, since <math>10+3\approx 12</math> in the pentagon with apparently obtuse angles. Plugging this back into equations <math>(1)</math> and <math>(2)</math> we find that <math>a = \frac{44}{3}</math> and <math>b= \frac{135}{10}=\frac{27}{2}</math>. | ||
+ | |||
+ | We desire <math>3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{216+88+81}{6}=\frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math> | ||
+ | |||
+ | ==Solution 3 (Ptolemy's but Quicker)== | ||
+ | |||
+ | Let us set <math>x</math> to be <math>AC=BD=CE</math> and <math>y</math> to be <math>BE</math> and <math>z</math> to be <math>AD</math>. It follow from applying [[Ptolemy's Theorem]] on <math>ABCD</math> to get <math>x^2=9+10z</math>. Applying Ptolemy's on <math>ACDE</math> gives <math>xz=42+10x</math>; and applying Ptolemy's on <math>BCDE</math> gives <math>x^2=100+3y</math>. So, we have the have the following system of equations: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | x^2 &= 9+10z \\ | ||
+ | x^2 &= 100+3y \\ | ||
+ | xz &= 42+10x | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | |||
+ | From <math>(3)</math>, we have <math>42=(z-10)x</math>. Isolating the x gives <math>x=\dfrac{42}{z-10}</math>. By setting <math>(1)</math> and <math>(2)</math> equal, we have <math>x^2=9+10z=100+3y</math>. Manipulating it gives <math>3y=10z-91</math>. Finally, plugging back into <math>(2)</math> gives <math>x^2=100+10z-91=10z+9</math>. Plugging in the <math>x=\dfrac{42}{z-10}</math> as well gives | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \left(\frac{42}{z-10}\right)^2 &= 10z+9\\ | ||
+ | 10z^3 - 191z^2 + 820z + 900 &= 1764\\ | ||
+ | 10z^3 - 191z^2 + 820z - 864 &= 0\\ | ||
+ | (5z-8)(2z-27)(z-4) &=0 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | It is impossible for <math>z<10</math> for <math>x<0</math>; that means <math>z=\frac{27}{2}</math>. That means <math>x = 12</math> and <math>y = \frac{44}{3}</math>. | ||
+ | |||
+ | Thus, the sum of all diagonals is <math>3x+y+z = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = 385/6</math>, which implies our answer is <math>m+n = 385+6 = \fbox{391 \textbf{(D)}}</math>. | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
+ | ==Solution 4== | ||
+ | [[File:2014AMC12BProblem24Solution4.png|center|400px]] | ||
+ | Let <math>BE = a</math>, <math>AC = CE = BD = b</math> | ||
+ | |||
+ | By [[Ptolemy's theorem]] for quadrilateral <math>ABCE</math>, <math>AB \cdot CE + BC \cdot AE = BE \cdot AC</math>, <math>3b + 140 = ab</math>, <math>a = 3 + \frac{140}{b}</math> | ||
+ | |||
+ | By [[Ptolemy's theorem]] for quadrilateral <math>BCDE</math>, <math>CD \cdot BE + BC \cdot DE = BD \cdot CE</math>, <math>3a + 100 = b^2</math> | ||
+ | |||
+ | <math>3(3 + \frac{140}{b}) + 100 = b^2</math>, <math>b^3 - 109 b -420 = 0</math>, <math>(b-12)(b+7)(b+5) = 0</math>, <math>b = 12</math> | ||
+ | |||
+ | <math>a = 3 + \frac{140}{12} = \frac{44}{3}</math> | ||
+ | |||
+ | By [[Ptolemy's theorem]] for quadrilateral <math>ABDE</math>, <math>AE \cdot BD + AB \cdot DE = AD \cdot BE</math>, <math>AD \cdot a = 14b + 30</math>, <math>AD = \frac{27}{2}</math> | ||
+ | |||
+ | <math>\frac{m}{n} = 12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, <math>385 + 6 = \boxed{\textbf{(D) }391}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:09, 1 January 2023
Contents
Problem
Let be a pentagon inscribed in a circle such that , , and . The sum of the lengths of all diagonals of is equal to , where and are relatively prime positive integers. What is ?
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=1-2vT_GIceA
Solution 1
Let , , and . Let be on such that . In we have . We use the Law of Cosines on to get . Eliminating we get which factorizes as Discarding the negative roots we have . Thus . For , we use Ptolemy's theorem on cyclic quadrilateral to get . For , we use Ptolemy's theorem on cyclic quadrilateral to get .
The sum of the lengths of the diagonals is so the answer is
Solution 2
Let denote the length of a diagonal opposite adjacent sides of length and , for sides and , and for sides and . Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:
Using equations and , we obtain:
and
Plugging into equation , we find that:
Or similarly into equation to check:
, being a length, must be positive, implying that . In fact, this is reasonable, since in the pentagon with apparently obtuse angles. Plugging this back into equations and we find that and .
We desire , so it follows that the answer is
Solution 3 (Ptolemy's but Quicker)
Let us set to be and to be and to be . It follow from applying Ptolemy's Theorem on to get . Applying Ptolemy's on gives ; and applying Ptolemy's on gives . So, we have the have the following system of equations:
From , we have . Isolating the x gives . By setting and equal, we have . Manipulating it gives . Finally, plugging back into gives . Plugging in the as well gives
It is impossible for for ; that means . That means and .
Thus, the sum of all diagonals is , which implies our answer is .
~sml1809
Solution 4
Let ,
By Ptolemy's theorem for quadrilateral , , ,
By Ptolemy's theorem for quadrilateral , ,
, , ,
By Ptolemy's theorem for quadrilateral , , ,
,
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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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