Difference between revisions of "2018 AMC 12A Problems/Problem 23"
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Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}</math> | Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}</math> | ||
− | Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math> | + | Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math>. |
~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) |
Revision as of 12:19, 22 June 2021
Contents
Problem
In
and
Points
and
lie on sides
and
respectively, so that
Let
and
be the midpoints of segments
and
respectively. What is the degree measure of the acute angle formed by lines
and
Solution 1
Let be the origin, and
lie on the x axis.
We can find and
Then, we have and
is the midpoint of
and
, or
Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.
This evaluates to
Now, using sum to product identities, we have this equal to
so the answer is
~lifeisgood03
Note: Though this solution is excellent, setting makes life a tad bit easier
~MathleteMA
Solution 2 (Overkill)
Note that , the midpoint of major arc
on
is the Miquel Point of
(Because
). Then, since
, this spiral similarity carries
to
. Thus, we have
, so
.
But, we have ; thus
.
Then, as is the midpoint of the major arc, it lies on the perpendicular bisector of
, so
. Since we want the acute angle, we have
, so the answer is
.
~stronto
Sidenote
For another way to find , note that
giving
as desired.
Solution 3 (Nice, I Think?)
Let the bisector of intersect
at
We have
so
We claim that
is parallel to this angle bisector, meaning that the acute angle formed by
and
is
meaning that the answer is
.
To prove this, let be the midpoint of
where
and
are the points on
and
respectively, such that
(The points given in this problem correspond to
but the idea we're getting at is that
will ultimately not matter.) Since
and
vary linearly with
the locus of all points
must be a line. Notice that
so
lies on this line. Let
be the intersection of this line with
(we know that this line will intersect
and not
because
). Notice that
Let Then
and
Thus,
By the Angle Bisector Theorem,
so
Since
is the midpoint of
we also have
Notice that:
Since the line containing all points
must be parallel to
This concludes the proof.
The critical insight to finding this solution is that the length probably shouldn't matter because a length ratio of
or
(as in the problem) is exceedingly unlikely to generate nice angles. This realization then motivates the idea of looking at all points similar to
which then leads to looking at the most convenient such point (in this case, the one that lies on
).
~sujaykazi
Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
Solution 4
Let the mid-point of be
and the mid-point of
be
.
Since
and
, we can conclude that
.
Similarly, we can conclude that
. Construct
and intersects
at
, which gives
.
Since
,
, we can find the value of
, which is equal to
. Since
, which means
, we can infer that
.
As we are required to give the acute angle formed, the final answer would be
, which is
.
~Surefire2019
Solution 5 (Simplest, I think?)
Link , extend
to
so that
. Then link
and
.
,
is the middle point of
and
is the middle line of
Notice that
As a result, ,
,
Also,
As a result,
Therefore,
Since we are asked for the acute angle between the two lines, the answer to this problem is .
~Solution by (Frank FYC)
Solution 6 (Olympiad Nuke)
By https://artofproblemsolving.com/community/c6h489748p2745891, we get that is parallel to the angle bisector of
Thus,
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc12a/473
~ dolphin7
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.