Difference between revisions of "2006 AIME I Problems/Problem 8"
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[[Hexagon]] <math> ABCDEF </math> is divided into five [[rhombus]]es, <math> \mathcal{P, Q, R, S,} </math> and <math> \mathcal{T,} </math> as shown. Rhombuses <math> \mathcal{P, Q, R,} </math> and <math> \mathcal{S} </math> are [[congruent (geometry) | congruent]], and each has [[area]] <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math> \mathcal{T}</math>. Given that <math> K </math> is a [[positive integer]], find the number of possible values for <math> K</math>. | [[Hexagon]] <math> ABCDEF </math> is divided into five [[rhombus]]es, <math> \mathcal{P, Q, R, S,} </math> and <math> \mathcal{T,} </math> as shown. Rhombuses <math> \mathcal{P, Q, R,} </math> and <math> \mathcal{S} </math> are [[congruent (geometry) | congruent]], and each has [[area]] <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math> \mathcal{T}</math>. Given that <math> K </math> is a [[positive integer]], find the number of possible values for <math> K</math>. | ||
− | + | <asy> | |
+ | // TheMathGuyd | ||
+ | size(8cm); | ||
+ | pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); | ||
+ | draw(A--B--C--D--EE--F--cycle); | ||
+ | draw(F--G--(2.1,0)); | ||
+ | draw(C--H--(2.1,0)); | ||
+ | draw(G--(2.1,-3.2)); | ||
+ | draw(H--(2.1,-3.2)); | ||
+ | label("$\mathcal{T}$",(2.1,-1.6)); | ||
+ | label("$\mathcal{P}$",(0,-1),NE); | ||
+ | label("$\mathcal{Q}$",(4.2,-1),NW); | ||
+ | label("$\mathcal{R}$",(0,-2.2),SE); | ||
+ | label("$\mathcal{S}$",(4.2,-2.2),SW); | ||
+ | </asy> | ||
− | == Solution == | + | == Solution 1== |
Let <math>x</math> denote the common side length of the rhombi. | Let <math>x</math> denote the common side length of the rhombi. | ||
Let <math>y</math> denote one of the smaller interior [[angle]]s of rhombus <math> \mathcal{P} </math>. Then <math>x^2\sin(y)=\sqrt{2006}</math>. We also see that <math>K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y</math>. Thus <math>K</math> can be any positive integer in the [[interval]] <math>(0, 2\sqrt{2006})</math>. | Let <math>y</math> denote one of the smaller interior [[angle]]s of rhombus <math> \mathcal{P} </math>. Then <math>x^2\sin(y)=\sqrt{2006}</math>. We also see that <math>K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y</math>. Thus <math>K</math> can be any positive integer in the [[interval]] <math>(0, 2\sqrt{2006})</math>. | ||
− | <math>2\sqrt{2006} = \sqrt{8024}</math> and <math>89^2 = 7921 < 8024 < 8100 = 90^2</math>, so <math>K</math> can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for <math>K</math> is 089. | + | <math>2\sqrt{2006} = \sqrt{8024}</math> and <math>89^2 = 7921 < 8024 < 8100 = 90^2</math>, so <math>K</math> can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for <math>K</math> is <math>\boxed{089}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where <math>w*h=\sqrt{2006}</math>. The height of rhombus T would be 2h, and the width would be <math>\sqrt{w^2-h^2}</math>. Substitute the first equation to get <math>\sqrt{\frac{2006}{h^2}-h^2}</math>. Then the area of the rhombus would be <math>2h * \sqrt{\frac{2006}{h^2}-h^2}</math>. Combine like terms to get <math>2 * \sqrt{2006-h^4}</math>. This expression equals an integer K. <math>2006-h^4</math> specifically must be in the form <math>n^2/4</math>. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of <math>n^2</math> for <math>2006-h^4</math>. Now, quick testing shows that <math>44^2 < 2006</math> and <math>45^2>2006</math>, but we must also test <math>44.5^2</math>, because the product of two will make it an integer. <math>44.5^2</math> is also less than <math>2006</math>, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us <math>44*2+1=</math> <math>\boxed{089}</math> | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
+ | |||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); | ||
+ | draw(A--B--C--D--EE--F--cycle); | ||
+ | label("$A$",A,2*N); | ||
+ | label("$B$",B,2*N); | ||
+ | label("$C$",C,2*E); | ||
+ | label("$D$",D,2*S); | ||
+ | label("$E$",EE,2*S); | ||
+ | label("$F$",F,2*W); | ||
+ | label("$G$",(0.47,-1.55),NW); | ||
+ | label("$H$",(3.73,-1.55),NE); | ||
+ | label("$I$",I,2*N); | ||
+ | label("$J$",J,2*S); | ||
+ | label("$K$",K,2*SW); | ||
+ | draw(F--C); | ||
+ | draw(F--G--(2.1,0)); | ||
+ | draw(C--H--(2.1,0)); | ||
+ | draw(G--(2.1,-3.2)); | ||
+ | draw(H--(2.1,-3.2)); | ||
+ | draw((2.1,0)--(2.1,-3.2)); | ||
+ | </asy> | ||
+ | To determine the possible values of <math>[GIHJ],</math> we must determine the maximum and minimum possible areas. | ||
+ | |||
+ | In the case where the <math>4</math> rhombi are squares, we have <math>[GIHJ]=0,</math> implying the minimum possible positive-integer-valued area is <math>1.</math> | ||
+ | |||
+ | Denote the length <math>HC=a</math> and <math>KH=b.</math> We have | ||
+ | <cmath>KI=\sqrt{a^2-b^2}</cmath> | ||
+ | by the Pythagorean Theorem, which implies | ||
+ | <cmath>[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}</cmath> | ||
+ | and | ||
+ | <cmath>[GIHJ]=2b\sqrt{a^2-b^2}.</cmath> | ||
+ | The first equation yields | ||
+ | <cmath>\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.</cmath> | ||
+ | Plugging into the second, we have | ||
+ | <cmath>[GIHJ]=2\sqrt{2006}\frac{b}{a}.</cmath> | ||
+ | The maximal value of <math>\frac{b}{a}</math> occurs when the height of <math>ABCDEF</math> is minimized, which means | ||
+ | <cmath>\frac{b}{a}\leq 1.</cmath> | ||
+ | Plugging back up, we have | ||
+ | <cmath>[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.</cmath> | ||
+ | We have | ||
+ | <cmath>\lfloor \sqrt{8024} \rfloor = 89,</cmath> | ||
+ | thus our answer is | ||
+ | <cmath>89-1+1=\boxed{089}.</cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=7|num-a=9}} | {{AIME box|year=2006|n=I|num-b=7|num-a=9}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
− | [[Category: | + | [[Category:Introductory Trigonometry Problems]] |
+ | {{MAA Notice}} |
Latest revision as of 15:05, 15 January 2024
Problem
Hexagon is divided into five rhombuses,
and
as shown. Rhombuses
and
are congruent, and each has area
Let
be the area of rhombus
. Given that
is a positive integer, find the number of possible values for
.
Solution 1
Let denote the common side length of the rhombi.
Let
denote one of the smaller interior angles of rhombus
. Then
. We also see that
. Thus
can be any positive integer in the interval
.
and
, so
can be any integer between 1 and 89, inclusive. Thus the number of positive values for
is
.
Solution 2
Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where . The height of rhombus T would be 2h, and the width would be
. Substitute the first equation to get
. Then the area of the rhombus would be
. Combine like terms to get
. This expression equals an integer K.
specifically must be in the form
. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of
for
. Now, quick testing shows that
and
, but we must also test
, because the product of two will make it an integer.
is also less than
, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us
-jackshi2006
Solution 3
To determine the possible values of
we must determine the maximum and minimum possible areas.
In the case where the rhombi are squares, we have
implying the minimum possible positive-integer-valued area is
Denote the length and
We have
by the Pythagorean Theorem, which implies
and
The first equation yields
Plugging into the second, we have
The maximal value of
occurs when the height of
is minimized, which means
Plugging back up, we have
We have
thus our answer is
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.