Difference between revisions of "2024 AMC 12A Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
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> | ||
Line 15: | Line 16: | ||
<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> |
Revision as of 22:10, 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 . Apply the Law of Cosines on :
Let . Apply the Law of Cosines on :
By Ptolemy’s Theorem, Since , The answer is .
(someone please insert diagram)
~lptoggled, formatting by eevee9406
Solution 2 (Law of Cosines + Law of Sines)
Draw diagonals and . By Law of Cosines,
~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.