2024 AMC 12A Problems/Problem 19
Contents
[hide]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
Solution 2 (Law of Cosines + Law of Sines)
Draw diagonals and
. By Law of Cosines,
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
Solution 3 (Law of Cosines + Cyclic Quadrilateral Property)
Draw diagonals and
. First, use Law of Cosines to get that
. Since
is cyclic,
, so Law of Cosines once again with respect to
on triangle
leads to
begin{align*}
9=5^2+7^2-2(7)(5)\cos\theta \
&= 74-70\cos\theta. \
\end{align*}
Solving yields
. Finally, in
, we have
\boxed{\textbf{(D) }\frac{39}{7}}$.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=f32mBtYTZp8
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 |
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