Difference between revisions of "2024 AMC 12A Problems/Problem 19"
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| − | == | + | ==Problem== |
| − | <math>\angle CBA=60 ^\circ</math> by | + | Cyclic quadrilateral <math>ABCD</math> has lengths <math>BC=CD=3</math> and <math>DA=5</math> with <math>\angle CDA=120^\circ</math>. What is the length of the shorter diagonal of <math>ABCD</math>? |
| − | Let <math>AC=u</math> | + | |
| − | <cmath>u^2=3^2+5^2-2(3)(5)cos120</cmath> | + | <math>\textbf{(A) }\frac{31}7 \qquad \textbf{(B) }\frac{33}7 \qquad \textbf{(C) }5 \qquad \textbf{(D) }\frac{39}7 \qquad \textbf{(E) }\frac{41}7 \qquad</math> |
| + | |||
| + | ==Solution 1== | ||
| + | First, <math>\angle CBA=60 ^\circ</math> by properties of cyclic quadrilaterals. | ||
| + | Let <math>AC=u</math>. We apply the [[Law of Cosines]] on <math>\triangle ACD</math>: | ||
| + | <cmath>u^2=3^2+5^2-2(3)(5)\cos120</cmath> | ||
<cmath>u=7</cmath> | <cmath>u=7</cmath> | ||
| − | Let <math>AB=v</math> | + | Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>: |
| − | <cmath>7^2=3^2+v^2-2(3)(v)cos60</cmath> | + | <cmath>7^2=3^2+v^2-2(3)(v)\cos60</cmath> |
<cmath>v=\frac{3\pm13}{2}</cmath> | <cmath>v=\frac{3\pm13}{2}</cmath> | ||
<cmath>v=8</cmath> | <cmath>v=8</cmath> | ||
| − | By | + | 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}<5</math> | Since <math>\frac{39}{7}<5</math> | ||
| − | The answer is <math>\ | + | The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>. |
| + | |||
| + | ~formatting by eevee9406 | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:07, 8 November 2024
Problem
Cyclic quadrilateral 
has lengths 
and 
with 
. What is the length of the shorter diagonal of ?
Solution 1
First, 
by properties of cyclic quadrilaterals.
Let 
. We apply the Law of Cosines on 
:
![\[u^2=3^2+5^2-2(3)(5)\cos120\]](http://latex.artofproblemsolving.com/a/6/1/a615672aa4ce2928a13cd3cf65f59e516a9d0281.png)
![\[u=7\]](http://latex.artofproblemsolving.com/e/c/7/ec7657c1ae5d5e39fe5611d3dbf210b65d39c41f.png)
Let 
. Apply the Law of Cosines on 
:
![\[7^2=3^2+v^2-2(3)(v)\cos60\]](http://latex.artofproblemsolving.com/4/b/4/4b4bb1df5d46ccf1a6822e0a34a641412c9514d7.png)
![\[v=\frac{3\pm13}{2}\]](http://latex.artofproblemsolving.com/e/f/d/efd53a5925c1959130cfb0ccb9b4c0545adfd1a6.png)
![\[v=8\]](http://latex.artofproblemsolving.com/5/c/8/5c821fed96435f71d8b2086f2ac5d7a733229173.png)
By Ptolemy’s Theorem,
![\[AB \cdot CD+AD \cdot BC=AC \cdot BD\]](http://latex.artofproblemsolving.com/8/f/9/8f90407491cc1fcb3e335391dabab417990ba5c0.png)
![\[8 \cdot 3+5 \cdot 3=7BD\]](http://latex.artofproblemsolving.com/9/4/9/9498c005ba103e2bb5e799ae3d3a05f0e615b88a.png)
![\[BD=\frac{39}{7}\]](http://latex.artofproblemsolving.com/c/5/1/c516c5649962b98d6b4f2fa26c9ebeba2eefde3a.png)
Since 
The answer is 
.
~formatting by eevee9406
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. 
