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Difference between revisions of "2006 AIME I Problems/Problem 8"
(→‎Solution 2)
 
(18 intermediate revisions by 10 users not shown)
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== Problem ==
 
== Problem ==
−
[[Hexagon]] <math> ABCDEF </math> is divided into four [[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));
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draw(C--H--(2.1,0));
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draw(G--(2.1,-3.2));
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draw(H--(2.1,-3.2));
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label("$\mathcal{T}$",(2.1,-1.6));
 +
label("$\mathcal{P}$",(0,-1),NE);
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label("$\mathcal{Q}$",(4.2,-1),NW);
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label("$\mathcal{R}$",(0,-2.2),SE);
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label("$\mathcal{S}$",(4.2,-2.2),SW);
 +
</asy>
  
−
{{image}}
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== 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>
  
  
−
== Solution ==
+
-jackshi2006
−
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>\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>.
+
==Solution 3==
−
<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.
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<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);
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label("$B$",B,2*N);
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label("$C$",C,2*E);
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label("$D$",D,2*S);
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label("$E$",EE,2*S);
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label("$F$",F,2*W);
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label("$G$",(0.47,-1.55),NW);
 +
label("$H$",(3.73,-1.55),NE);
 +
label("$I$",I,2*N);
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label("$J$",J,2*S);
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label("$K$",K,2*SW);
 +
draw(F--C);
 +
draw(F--G--(2.1,0));
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draw(C--H--(2.1,0));
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draw(G--(2.1,-3.2));
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draw(H--(2.1,-3.2));
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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
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<cmath>KI=\sqrt{a^2-b^2}</cmath>
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by the Pythagorean Theorem, which implies
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<cmath>[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}</cmath>
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and
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<cmath>[GIHJ]=2b\sqrt{a^2-b^2}.</cmath>
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The first equation yields
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<cmath>\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.</cmath>
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Plugging into the second, we have
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<cmath>[GIHJ]=2\sqrt{2006}\frac{b}{a}.</cmath>
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The maximal value of <math>\frac{b}{a}</math> occurs when the height of <math>ABCDEF</math> is minimized, which means
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<cmath>\frac{b}{a}\leq 1.</cmath>
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Plugging back up, we have
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<cmath>[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.</cmath>
 +
We have
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<cmath>\lfloor \sqrt{8024} \rfloor = 89,</cmath>
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thus our answer is
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<cmath>89-1+1=\boxed{089}.</cmath>
 
== See also ==
 
== See also ==
−
* [[2006 AIME I Problems/Problem 7 | Previous problem]]
+
{{AIME box|year=2006|n=I|num-b=7|num-a=9}}
−
* [[2006 AIME I Problems/Problem 9 | Next problem]]
+
 
−
* [[2006 AIME I Problems]]
 
 
[[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)
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Problem 9
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