Difference between revisions of "2020 AIME II Problems/Problem 4"
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==Solution 3== | ==Solution 3== | ||
− | A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math> | + | A <math>90^\circ</math> degree rotation is obvious. Let's look at <math>C</math> and <math>C'</math>. They are very close to each other. Let's join <math>C</math> and <math>C'</math> with a line. Then construct a perpendicular bisector to <math>\overline{CC'}</math> with the midpoint being <math>M</math> which is at <math>(20, 1)</math>. We also draw a point <math>N</math> on the perpendicular bisector such that <math>\angle CNC'</math> is <math>90^\circ</math>. That point <math>N</math> is the same distance to <math>M</math> as <math>M</math> is to <math>C</math> but it is on a line perpendicular to <math>\overline{CM}</math> Therefore <math>N</math> is at <math>(20+1, 1-4)</math><math>. The sum is </math>90+20+1+1-4=108$. |
==Video Solution== | ==Video Solution== |
Revision as of 14:30, 8 June 2020
Problem
Triangles and
lie in the coordinate plane with vertices
,
,
,
,
,
. A rotation of
degrees clockwise around the point
where
, will transform
to
. Find
.
Solution
After sketching, it is clear a rotation is done about
. Looking between
and
,
and
. Solving gives
. Thus
.
~mn28407
Solution 2 (Official MAA)
Because the rotation sends the vertical segment to the horizontal segment
, the angle of rotation is
degrees clockwise. For any point
not at the origin, the line segments from
to
and from
to
are perpendicular and are the same length. Thus a
clockwise rotation around the point
sends the point
to the point
. This has the solution
. The requested sum is
.
Solution 3
A degree rotation is obvious. Let's look at
and
. They are very close to each other. Let's join
and
with a line. Then construct a perpendicular bisector to
with the midpoint being
which is at
. We also draw a point
on the perpendicular bisector such that
is
. That point
is the same distance to
as
is to
but it is on a line perpendicular to
Therefore
is at
90+20+1+1-4=108$.
Video Solution
~IceMatrix
See Also=
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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