Difference between revisions of "2021 AMC 12B Problems/Problem 24"
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<math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | <math>\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113</math> | ||
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==Solution 1== | ==Solution 1== | ||
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Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>. Remark that by symmetry <math>X</math> is the midpoint of both <math>\overline{PQ}</math> and <math>\overline{RS}</math>, so <math>XP = XQ = 3</math> and <math>XR = XS = 4</math>. Now note that since <math>\angle APB = \angle ARB = 90^\circ</math>, quadrilateral <math>ARPB</math> is cyclic, and so | ||
<cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>. | <cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34</math>. | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
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+ | ==Video Solution (Simple: Using trigonometry and Equations)== | ||
+ | https://youtu.be/ZB-VN02H6mU | ||
+ | ~hippopotamus1 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2021|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:10, 17 May 2021
Contents
Problem
Let be a parallelogram with area
. Points
and
are the projections of
and
respectively, onto the line
and points
and
are the projections of
and
respectively, onto the line
See the figure, which also shows the relative locations of these points.
Suppose and
and let
denote the length of
the longer diagonal of
Then
can be written in the form
where
and
are positive integers and
is not divisible by the square of any prime. What is
Solution 1
Let denote the intersection point of the diagonals
and
. Remark that by symmetry
is the midpoint of both
and
, so
and
. Now note that since
, quadrilateral
is cyclic, and so
which implies
.
Thus let be such that
and
. Then Pythagorean Theorem on
yields
, and so
Solving this for
yields
, and so
The requested answer is
.
Solution 2(Trig)
Let denote the intersection point of the diagonals
and
and let
. Then, by the given conditions,
. So,
Combining the above 3 equations, we get
Since we want to find
we let
Then
Solving this, we get
so
.
Solution 3 (Similar Triangles and Algebra)
Let be the intersection of diagonals
and
. By symmetry
,
and
, so now we have reduced all of the conditions one quadrant. Let
.
,
by similar triangles and using the area condition we get
. Note that it suffices to find
because we can double and square it to get
. Solving for
in the above equation, and then using
.
Solution 4 (Similar Triangles)
Again, Let be the intersection of diagonals
and
. Note that triangles
and
are similar because they are right triangles and share
. First, call the length of
. By the definition of an area of a parallelogram,
, so
. Using similar triangles on
and
,
. Therefore, finding
,
. Now, applying the Pythagorean theorem once, we find
+
=
. Solving this equation for
, we find
.
Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
~ pi_is_3.14
Video Solution (Simple: Using trigonometry and Equations)
https://youtu.be/ZB-VN02H6mU ~hippopotamus1
See Also
2021 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.