Difference between revisions of "2021 AMC 12B Problems/Problem 24"
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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{ | + | <cmath>XR\cdot XA = XP\cdot XB,</cmath>which implies <math>\tfrac{XB}{XA} = \tfrac{XP}{XR} = \tfrac34</math>. |
Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>. | Thus let <math>x> 0</math> be such that <math>XA = 3x</math> and <math>XB = 4x</math>. Then Pythagorean Theorem on <math>\triangle APX</math> yields <math>AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.</cmath>Solving this for <math>x^2</math> yields <math>x^2 = \tfrac12 + \tfrac{\sqrt{41}}8</math>, and so<cmath>(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.</cmath>The requested answer is <math>32 + 8 + 41 = \boxed{81}</math>. | ||
+ | |||
==Solution 2(Trig) == | ==Solution 2(Trig) == | ||
Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>, and let <math>\theta = \angle{COB}</math>. Then, by the given conditions, <math>XR = 4</math>, <math>XQ = 3</math>, <math>[XCB] = \frac{15}{4}</math>. So, | Let <math>X</math> denote the intersection point of the diagonals <math>AC</math> and <math>BD</math>, and let <math>\theta = \angle{COB}</math>. Then, by the given conditions, <math>XR = 4</math>, <math>XQ = 3</math>, <math>[XCB] = \frac{15}{4}</math>. So, |
Revision as of 22:36, 15 February 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
simple Video Solution Using trigonometry and Equations
https://youtu.be/ZB-VN02H6mU ~hippopotamus1
Solution
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,
Combine the above 3 equations we get
.
Since we want to find
, let
, then
.
Solve, we get
, so
.
~sequoia
Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
~ pi_is_3.14
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.