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> | ||
− | == | + | ==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>. | ||
− | 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} | + | 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}=15</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{\textbf{(A)} ~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, | ||
<cmath> XC = \frac{3}{\cos \theta}</cmath> | <cmath> XC = \frac{3}{\cos \theta}</cmath> | ||
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Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},</math> we let <math>x = \frac{1}{\cos^2 \theta}.</math> Then | Since we want to find <math>d^2 = 4XB^2 = \frac{64}{\cos^2 \theta},</math> we let <math>x = \frac{1}{\cos^2 \theta}.</math> Then | ||
<cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | <cmath> \frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}.</cmath> | ||
− | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8},</math> so <math>d^2 = 64x = 32 + 8\sqrt{41} | + | Solving this, we get <math>x = \frac{4 + \sqrt{41}}{8},</math> so <math>d^2 = 64x = 32 + 8\sqrt{41} \rightarrow 32+8+41=\boxed{\textbf{(A)} ~81}</math> |
==Solution 3 (Similar Triangles and Algebra)== | ==Solution 3 (Similar Triangles and Algebra)== | ||
− | Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. By symmetry <math>[\triangle XCB] = \frac{15}{4}</math>, <math>XQ = 3</math> and <math>XR = 4</math>, so now we have reduced all of the conditions one quadrant. Let <math>CQ = x</math>. <math>XC = \sqrt{x^2+9}</math>, <math>RB = \frac{4x}{3}</math> by similar triangles and using the area condition we get <math>\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}</math>. Note that it suffices to find <math> | + | Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. By symmetry <math>[\triangle XCB] = \frac{15}{4}</math>, <math>XQ = 3</math> and <math>XR = 4</math>, so now we have reduced all of the conditions one quadrant. Let <math>CQ = x</math>. <math>XC = \sqrt{x^2+9}</math>, <math>RB = \frac{4x}{3}</math> by similar triangles and using the area condition we get <math>\frac{4}{3} \cdot x \cdot \sqrt{x^2+9} = \frac{15}{2}</math>. Note that it suffices to find <math>XB = \frac{4}{3}\sqrt{x^2+9}</math> because we can double and square it to get <math>d^2</math>. Solving for <math>a = x^2</math> in the above equation, and then using <math>d^2 = \frac{64}{9}(x^2+9) = 8\sqrt{41} + 32 \Rightarrow 8+41+32=\boxed{\textbf{(A)} ~81}</math> |
+ | |||
+ | ==Solution 4 (Similar Triangles)== | ||
+ | Again, Let <math>X</math> be the intersection of diagonals <math>AC</math> and <math>BD</math>. Note that triangles <math>\triangle QXC</math> and <math>\triangle BXR</math> are similar because they are right triangles and share <math>\angle CXQ</math>. First, call the length of <math>XB = \frac{d}{2}</math>. By the definition of an area of a parallelogram, <math>CQ \cdot 2XB = 15</math>, so <math>CQ = \frac{15}{d}</math>. Using similar triangles on <math>\triangle QXC</math> and <math>\triangle BXR</math>, <math>\frac{CQ}{XQ} = \frac{BR}{XR}</math>. Therefore, finding <math>BR</math>, <math>BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}</math>. Now, applying the Pythagorean theorem once, we find <math>(\frac{20}{d}) ^2</math> + <math>(4)^2</math> = <math>(\frac{d}{2}) ^2</math>. Solving this equation for <math>d^2</math>, we find <math>d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \rightarrow 32+8+41= \boxed{\textbf{(A)} ~81}</math> | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>BQ = PD = x.</math> We know that the area of the parallelogram is <math>15,</math> so it follows that <math>[\triangle{BCD}] = [\triangle{BAD}] = \tfrac{15}{2}</math> and the height of each triangle, which are also the lengths of <math>QC</math> and <math>AP,</math> is <math>\tfrac{15}{2(x+3)}.</math> Suppose that <math>E = RS \cap BD.</math> Because <math>\angle{BRE} = \angle{CQE}</math> and <math>\angle{BER} = \angle{CQD},</math> we have <math>\triangle{BRE} \sim \triangle{CQE}.</math> The length of <math>CE,</math> by the Pythagorean Theorem is <math>\sqrt{3^2+(\tfrac{15}{2(x+3)})^2}</math> and the length of <math>BR,</math> by the Pythagorean Theorem on <math>\triangle{BRE},</math> is <math>\sqrt{(x+3)^2 - 4}.</math> Note that | ||
+ | <cmath> \sin{\angle QEC} = \frac{CQ}{CE} = \frac{BR}{BE} </cmath> | ||
+ | Substituting in our values, | ||
+ | <cmath> \frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}</cmath> | ||
+ | To rid unnecessary computation, we let <math>(x+3)^2 = a.</math> The equation simplifies, after cross multiplying, to | ||
+ | <cmath> \sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2} </cmath> | ||
+ | <cmath> 36a^2 - 576a - 15^2\cdot 16 = 0</cmath> | ||
+ | <cmath> a^2-16a-100 =0 </cmath> | ||
+ | By the quadratic formula, <math>a \in \{\tfrac{16 - \sqrt{656}}{2}, \tfrac{16 + \sqrt{656}}{2}\},</math> so we discard the negative solution. The value of <math>BD^2</math> is | ||
+ | <cmath> BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}</cmath> | ||
+ | and the desired answer is <math>32+8+41 = \boxed{\textbf{(A)} ~81}</math> ~skyscraper | ||
+ | |||
+ | ==Solution 6 (Similar Triangles & Pythagorean Theorem)== | ||
+ | |||
+ | Let the intersection of <math>RS</math> and <math>BD</math> be <math>X</math>. | ||
+ | |||
+ | <math>\because</math> <math>\angle APX = \angle DSX</math> and <math>\angle AXP = \angle DXS</math>, <math>\triangle APX \sim \triangle DSX</math> by <math>AA</math> | ||
+ | |||
+ | <math>\therefore</math> <math>\frac{PA}{DS} = \frac68 = \frac34</math>, <math>DS = \frac43 \cdot PA</math> | ||
+ | |||
+ | By the Pythagorean theorem and the property of projection, <math>BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64</math>, <math>\frac{64}{9} \cdot PA^2 = BD^2 - 64</math> | ||
+ | |||
+ | <math>\because [ABCD] = PA \cdot BD = 15</math>, <math>\therefore PA = \frac{15}{BD}</math> | ||
+ | |||
+ | <cmath>\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64</cmath> | ||
+ | |||
+ | <cmath>\frac{1600}{BD^2} = BD^2 - 64</cmath> | ||
+ | |||
+ | <cmath>BD^4 - 64 BD^2 - 1600 = 0</cmath> | ||
+ | |||
+ | <cmath>BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}</cmath> | ||
+ | |||
+ | Therefore, the answer is <math>32 + 8 + 41 = \boxed{\textbf{(A) } 81}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=xtYSPxOMZlk | ||
== Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) == | == Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point) == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==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 12:07, 28 October 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Trig)
- 4 Solution 3 (Similar Triangles and Algebra)
- 5 Solution 4 (Similar Triangles)
- 6 Solution 5
- 7 Solution 6 (Similar Triangles & Pythagorean Theorem)
- 8 Video Solution by MOP 2024
- 9 Video Solution by OmegaLearn (Cyclic Quadrilateral and Power of a Point)
- 10 Video Solution (Simple: Using trigonometry and Equations)
- 11 See Also
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
Solution 5
Let We know that the area of the parallelogram is
so it follows that
and the height of each triangle, which are also the lengths of
and
is
Suppose that
Because
and
we have
The length of
by the Pythagorean Theorem is
and the length of
by the Pythagorean Theorem on
is
Note that
Substituting in our values,
To rid unnecessary computation, we let
The equation simplifies, after cross multiplying, to
By the quadratic formula,
so we discard the negative solution. The value of
is
and the desired answer is
~skyscraper
Solution 6 (Similar Triangles & Pythagorean Theorem)
Let the intersection of and
be
.
and
,
by
,
By the Pythagorean theorem and the property of projection, ,
,
Therefore, the answer is .
Video Solution by MOP 2024
https://youtube.com/watch?v=xtYSPxOMZlk
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.