Difference between revisions of "2024 AMC 12A Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
+ | |||
+ | <asy> | ||
+ | import geometry; | ||
+ | |||
+ | size(200); | ||
+ | |||
+ | // from geogebra lol | ||
+ | pair A = (-1.66, 0.33); | ||
+ | pair B = (-9.61277, 1.19799); | ||
+ | pair C = (-7.83974, 3.61798); | ||
+ | pair D = (-4.88713, 4.14911); | ||
+ | |||
+ | draw(circumcircle(A, B, C)); | ||
+ | |||
+ | draw(A--C); | ||
+ | draw(A--D); | ||
+ | draw(C--D); | ||
+ | draw(B--C); | ||
+ | draw(A--B); | ||
+ | |||
+ | label("$A$", A, E); | ||
+ | label("$B$", B, W); | ||
+ | label("$C$", C, NW); | ||
+ | label("$D$", D, N); | ||
+ | |||
+ | label("$7$", midpoint(A--C), SW); | ||
+ | label("$5$", midpoint(A--D), NE); | ||
+ | label("$3$", midpoint(C--D)+ dir(135)*0.3, N); | ||
+ | label("$3$", midpoint(B--C)+dir(180)*0.3, NW); | ||
+ | label("$8$", midpoint(A--B), S); | ||
+ | |||
+ | markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10); | ||
+ | markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10); | ||
+ | </asy> | ||
+ | ~diagram by erics118 | ||
+ | |||
First, <math>\angle CBA=60 ^\circ</math> by properties of cyclic quadrilaterals. | First, <math>\angle CBA=60 ^\circ</math> by properties of cyclic quadrilaterals. | ||
− | Let <math>AC=u</math>. | + | |
+ | Let <math>AC=u</math>. Apply the [[Law of Cosines]] on <math>\triangle ACD</math>: | ||
<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath> | <cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath> | ||
<cmath>u=7</cmath> | <cmath>u=7</cmath> | ||
− | |||
Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>: | Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>: | ||
Line 16: | Line 52: | ||
<cmath>v=8</cmath> | <cmath>v=8</cmath> | ||
− | + | By [[Ptolemy’s Theorem]], | |
− | By Ptolemy’s Theorem, | ||
<cmath>AB \cdot CD+AD \cdot BC=AC \cdot BD</cmath> | <cmath>AB \cdot CD+AD \cdot BC=AC \cdot BD</cmath> | ||
<cmath>8 \cdot 3+5 \cdot 3=7BD</cmath> | <cmath>8 \cdot 3+5 \cdot 3=7BD</cmath> | ||
<cmath>BD=\frac{39}{7}</cmath> | <cmath>BD=\frac{39}{7}</cmath> | ||
− | Since <math>\frac{39}{7}< | + | Since <math>\frac{39}{7}<7</math>, |
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>. | The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>. | ||
− | ~formatting by eevee9406 | + | |
+ | ~lptoggled, formatting by eevee9406, typo fixed by meh494, image by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Solution 2 (Law of Cosines + Law of Sines)== | ||
+ | Draw diagonals <math>AC</math> and <math>BD</math>. By Law of Cosines, | ||
+ | \begin{align*} | ||
+ | AC^2&=3^2 + 5^2 - 2(3)(5)\cos \left(\frac{2\pi}{3} \right) \\ | ||
+ | &= 9+25 +15 \\ | ||
+ | &=49. | ||
+ | \end{align*} | ||
+ | Since <math>AC</math> is positive, taking the square root gives <math>AC=7.</math> Let <math>\angle BDC=\angle CBD=x</math>. Since <math>\triangle BCD</math> is isosceles, we have <math>\angle BCD=180-2x</math>. Notice we can eventually solve <math>BD</math> using the Extended Law of Sines: <cmath>\frac{BD}{\sin(180-2x)}=2r,</cmath> where <math>r</math> is the radius of the circumcircle <math>ABCD</math>. Since <math>\sin(180-2x)=\sin(2x)=2\sin(x)\cos(x)</math>, we simply our equation: <cmath>\frac{BD}{2\sin(x)\cos(x)}=2r.</cmath> | ||
+ | Now we just have to find <math>\sin(x), \cos(x),</math> and <math>2r</math>. Since <math>ABCD</math> is cyclic, we have <math>\angle CBD = \angle CAD = x</math>. By Law of Cosines on <math>\triangle ADC</math>, we have <cmath>3^2=7^2 + 5^2 - 70\cos(x).</cmath> Thus, <math>\cos(x)=\frac{13}{14}.</math> Similarly, by Law of Sines on <math>\triangle ACD</math>, we have <cmath>\frac{7}{\sin\left(\frac{2\pi}{3} \right)}=2r.</cmath> Hence, <math>2r=\frac{14\sqrt3}{3}</math>. Now, using Law of Sines on <math>\triangle BCD</math>, we have <math>\frac{3}{\sin(x)}=2r= \frac{14\sqrt3}{3},</math> so <math>\sin(x)=\frac{3\sqrt3}{14}.</math> Therefore, <cmath>\frac{BD}{2\left(\frac{3\sqrt3}{14}\right) \left(\frac{13}{14} \right)}=\frac{14\sqrt3}{3}.</cmath> Solving, <math>BD = \frac{39}{7},</math> so the answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>. | ||
+ | |||
+ | ~evanhliu2009 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:38, 9 November 2024
Problem
Cyclic quadrilateral has lengths and with . What is the length of the shorter diagonal of ?
Solution 1
~diagram by erics118
First, by properties of cyclic quadrilaterals.
Let . Apply the Law of Cosines on :
Let . Apply the Law of Cosines on :
By Ptolemy’s Theorem, Since , The answer is .
~lptoggled, formatting by eevee9406, typo fixed by meh494, image by ~luckuso
Solution 2 (Law of Cosines + Law of Sines)
Draw diagonals and . By Law of Cosines, \begin{align*} AC^2&=3^2 + 5^2 - 2(3)(5)\cos \left(\frac{2\pi}{3} \right) \\ &= 9+25 +15 \\ &=49. \end{align*} Since is positive, taking the square root gives Let . Since is isosceles, we have . Notice we can eventually solve using the Extended Law of Sines: where is the radius of the circumcircle . Since , we simply our equation: Now we just have to find and . Since is cyclic, we have . By Law of Cosines on , we have Thus, Similarly, by Law of Sines on , we have Hence, . Now, using Law of Sines on , we have so Therefore, Solving, so the answer is .
~evanhliu2009
See also
2024 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.