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== | + | == Solutions == |
+ | === 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>. | ||
− | <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> | ||
== 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}} |
Revision as of 09:58, 25 September 2021
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 ?
Solutions
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
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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