Difference between revisions of "2018 AMC 12A Problems/Problem 23"
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− | + | Let the bisector of <math>\angle ATP</math> intersect <math>PA</math> at <math>X.</math> We have <math>\angle ATX = \angle PTX = 44^{\circ},</math> so <math>\angle TXA = 80^{\circ}.</math> We claim that <math>MN</math> is parallel to this angle bisector, meaning that the acute angle formed by <math>MN</math> and <math>PA</math> is <math>80^{\circ},</math> meaning that the answer is <math>\boxed{\textbf{(E)}}</math>. | |
To prove this, let <math>N(x)</math> be the midpoint of <math>U(x)G(x),</math> where <math>U(x)</math> and <math>G(x)</math> are the points on <math>PT</math> and <math>AT,</math> respectively, such that <math>PU = AG = x.</math> (The points given in this problem correspond to <math>x=1,</math> but the idea we're getting at is that <math>x</math> will ultimately not matter.) | To prove this, let <math>N(x)</math> be the midpoint of <math>U(x)G(x),</math> where <math>U(x)</math> and <math>G(x)</math> are the points on <math>PT</math> and <math>AT,</math> respectively, such that <math>PU = AG = x.</math> (The points given in this problem correspond to <math>x=1,</math> but the idea we're getting at is that <math>x</math> will ultimately not matter.) |
Revision as of 00:31, 15 February 2018
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
Let be the origin, and
lie on the x axis.
We can find and
Then, we have and
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)
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)
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.)
The critical insight to finding this solution is that the length probably shouldn't matter, because a length ratio of
or
(which show up in the problem) is exceedingly unlikely to generate nice angles. This idea 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!
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.