Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
R00tsofunity (talk | contribs) (combined solutions #1 and #2 into one.) |
R00tsofunity (talk | contribs) |
||
Line 22: | Line 22: | ||
===Solution 1.1 (Stewart's theorem)=== | ===Solution 1.1 (Stewart's theorem)=== | ||
− | Applying [[Stewart's theorem]] on <math>\triangle | + | Applying [[Stewart's theorem]] on <math>\triangle ABC</math> with cevian <math>\overline{CF}</math> using the directed lengths <math>AF = AC = 20</math> and <math>FB = 11-20 = -9</math>, we obtain <cmath>\begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*}</cmath> so <math>CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{\textbf{(C) }30}</math>. |
===Solution 1.2 (Double Cosine Law)=== | ===Solution 1.2 (Double Cosine Law)=== |
Revision as of 16:38, 5 January 2023
Contents
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
?
Olympiad Solution using Spiral Similarity
Construct the -antipode,
. Notice
by spiral similarity at
, thus
. Let
; by symmetry
as well and
from Law of Cosines in
, so by Law of Cosines in
we have
from which
. Now,
.
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,
Two solution methods follow from here.
Solution 1.1 (Stewart's theorem)
Applying Stewart's theorem on with cevian
using the directed lengths
and
, we obtain
so
.
Solution 1.2 (Double Cosine Law)
Note that so we may plug into Law of Cosines to find the angle's cosine:
So, we observe that we can use Law of Cosines again to find :
both ways.
- Kevinmathz
Solution 2
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)
Solution 4
Denote and
. Note that by Heron's formula the area of
is
so the
-coordinate of
(height of
above the
-axis) is easily computed by the base-height formula as
.
Now, since , the
-coordinate of
satisfies
and solving gives
.
The circumcircle of has radius
. We know by the perpendicular bisector rule that the circumcenter
is located directly below the midpoint of
(
-coordinate
).
So, the negative coordinate of
satisfies
and solving gives
.
It's also clear that point is going to be located directly below
on the circle, because the angle bisector intersects the circumcircle at the midpoint of the arc (Fact 5). Since the radius of the circle is
, we have the coordinates of
Solving for point (the point on the
-axis between
and
), we get that
.
So now we know six of the critical points: ;
;
;
;
;
.
We are now ready to add in the circumcircle of , which has radius
. From the above information,
,
, and
.
After a bit of simplification we end up with and
.
For the area of , the altitude dropped from vertex
has height
, and the base
has length
, so its area is
.
Thus, which after tons of cancellations becomes
.
We know from the perpendicular bisector rule that the circumcenter of
is located directly below the midpoint of
(
-coordinate
).
So, the negative -coordinate of
satisfies
, and solving gives
. Thus, the equation of the circumcircle of
is
.
Point is the intersection of this circle and the line
, which has equation
. So, we substitute
into the equation of the circle to get
.
After simplifying, we have (the
's cancel out), whose solutions are
and
. The first corresponds to the origin, and the second corresponds to point
. Thus the coordinates of
are
.
The coordinates of are
, so
Video Solution by Power of Logic(Trig and Power of a point)
~math2718281828459
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.