Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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− | Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that | + | Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem. |
Revision as of 19:48, 18 December 2016
Problem
Let be a triangle where
is the midpoint of
, and
is the angle bisector of
with
on
. Let
be the intersection of the median
and the bisector
. In addition
is equilateral with
. What is
?
Solution
Let and
. From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem.
is the midpoint of side
.
This implies that . Given that angle
is
degrees and angle
is
degrees, we can use the area formula to get
So, .....(1)
is angle bisector.
In the triangle , one has
, therefore
.....(2)
Furthermore, triangle is similar to triangle
, so
, therefore
....(3)
By (2) and (3) and the subtraction law of ratios, we get
Therefore , or
. So
.
Finally, using the law of cosine for triangle , we get
Solution 2 (Analytic)
![2013 AMC 12B 24.jpg](https://wiki-images.artofproblemsolving.com//a/ab/2013_AMC_12B_24.jpg)
Let us dilate triangle so that the sides of equilateral triangle
are all equal to
The purpose of this is to ease the calculations we make in the problem. Given this, we aim to find the length of segment
so that we can un-dilate triangle
by dividing each of its sides by
. Doing so will make it so that
, as desired, and doing so will allow us to get the length of
, whose square is our final answer.
Let the foot of the altitude from
to
On the coordinate plane, position
at
, and make
lie on the x-axis. Since points
,
, and
, are collinear,
must also lie on the x-axis. Additionally, since
,
, meaning that we can position point
at
. Now, notice that line
has the equation
and that line
has the equation
because angles
and
are both
. We can then position
at point
and
at point
. Quickly note that, because
is an angle bisector,
must pass through the point
.
We proceed to construct a system of equations. First observe that the midpoint of
must lie on
, with the equation
. The coordinates of
are
, and we can plug in these coordinates into the equation of line
, yielding that
For our second equation, notice that line
has equation
. Midpoint
must also lie on this line, and we can substitute coordinates again to get
Setting both equations equal to each other and multiplying both sides by , we have that
, which in turn simplifies into
when dividing the entire equation by
Using the quadratic formula, we have that
Here, we discard the positive root since
must lie to the left of the y-axis. Then, the coordinates of
are
, and the coordinates of
are
Seeing that segment
has half the length of side
, we have that the length of
is
Now, we divide each side length of by
, and from this,
will equal
See also
2013 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.