Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Stewart Bash)) |
MRENTHUSIASM (talk | contribs) m (→Solution 3) |
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\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
For <math>BF</math>, we have | For <math>BF</math>, we have | ||
<cmath> | <cmath> | ||
Line 69: | Line 68: | ||
\] | \] | ||
</cmath> | </cmath> | ||
− | |||
Hence, <math>BD = \frac{24 \cdot 11}{31}</math>. | Hence, <math>BD = \frac{24 \cdot 11}{31}</math>. | ||
Line 86: | Line 84: | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
Hence, <math>\sin \angle ABC = \frac{5 \sqrt{7}}{16}</math> and <math>\sin \angle BCA = \frac{11 \sqrt{7}}{64}</math>. | Hence, <math>\sin \angle ABC = \frac{5 \sqrt{7}}{16}</math> and <math>\sin \angle BCA = \frac{11 \sqrt{7}}{64}</math>. | ||
Hence, <math>\cot \angle BCA = \frac{57}{11 \sqrt{7}}</math>. | Hence, <math>\cot \angle BCA = \frac{57}{11 \sqrt{7}}</math>. | ||
Line 95: | Line 92: | ||
Now, we can compute <math>CF</math> whose expression is given in Equation (1). | Now, we can compute <math>CF</math> whose expression is given in Equation (1). | ||
We have <math>CF = 30</math>. | We have <math>CF = 30</math>. | ||
− | |||
Therefore, the answer is <math>\boxed{\textbf{(C) }30}</math>. | Therefore, the answer is <math>\boxed{\textbf{(C) }30}</math>. |
Revision as of 01:18, 28 November 2021
Problem
Triangle has side lengths
, and
. The bisector of
intersects
in point
, and intersects the circumcircle of
in point
. The circumcircle of
intersects the line
in points
and
. What is
?
Solution 1
Claim:
Proof: Note that and
meaning that our claim is true by AA similarity.
Because of this similarity, we have that by Power of a Point. Thus,
Now, note that and plug into Law of Cosines to find the angle's cosine:
So, we observe that we can use Law of Cosines again to find :
- kevinmathz
Solution 2
By the Inscribed Angle Theorem and the definition of angle bisectors note thatso
. Therefore
. By PoP, we can also express
as
so
and
. Let
. Applying Stewart’s theorem on
with cevian
we have
~Punxsutawney Phil
Solution 3
This solution is based on this figure: Image:2021_AMC_12B_(Nov)_Problem_24,_sol.png
Denote by the circumcenter of
.
Denote by
the circumradius of
.
In , following from the law of cosines, we have
For
, we have
The fourth equality follows from the property that
,
,
are concyclic.
The fifth and the ninth equalities follow from the property that
,
,
,
are concyclic.
Because bisects
, following from the angle bisector theorem, we have
Hence,
.
In , following from the law of cosines, we have
and
Hence,
and
.
Hence,
.
Now, we are ready to compute whose expression is given in Equation (2).
We get
.
Now, we can compute whose expression is given in Equation (1).
We have
.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.