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

Revision as of 03:20, 4 January 2023

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 $n^2/2$ . 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 $89-1+1=\boxed{089}.$

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