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Difference between revisions of "2006 AIME I Problems/Problem 8"

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== Problem ==
 
== Problem ==
 +
[[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 1==
 +
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>.
 +
<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>
  
−
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
 
  
−
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
+
-jackshi2006
  
−
== Solution ==
+
==Solution 3==
−
Let <math>x</math> denote the common side length of the rhombi.
+
<asy>
−
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>\displaystyle 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>.
+
size(8cm);
−
<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.
+
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>
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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:Intermediate Trigonometry Problems]]
+
[[Category:Introductory Trigonometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:05, 15 January 2024

Problem

Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$.

[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 1

Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2\sin(y)=\sqrt{2006}$. We also see that $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$. $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, so $K$ can be any integer between 1 and 89, inclusive. Thus the number of positive values for $K$ is $\boxed{089}$.


Solution 2

Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where $w*h=\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$. Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \sqrt{2006-h^4}$. This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/4$. 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 $n^2$ for $2006-h^4$. Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$, but we must also test $44.5^2$, because the product of two will make it an integer. $44.5^2$ is also less than $2006$, 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 $44*2+1=$ $\boxed{089}$


-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 $[GIHJ],$ we must determine the maximum and minimum possible areas.

In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$

Denote the length $HC=a$ and $KH=b.$ We have \[KI=\sqrt{a^2-b^2}\] by the Pythagorean Theorem, which implies \[[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}\] and \[[GIHJ]=2b\sqrt{a^2-b^2}.\] The first equation yields \[\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.\] Plugging into the second, we have \[[GIHJ]=2\sqrt{2006}\frac{b}{a}.\] The maximal value of $\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means \[\frac{b}{a}\leq 1.\] Plugging back up, we have \[[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.\] We have \[\lfloor \sqrt{8024} \rfloor = 89,\] thus our answer is \[89-1+1=\boxed{089}.\]

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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