2007 iTest Problems/Problem 32

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Problem

When a rectangle frames a parabola such that a side of the rectangle is parallel to the parabola's axis of symmetry, the parabola divides the rectangle into regions whose areas are in the ratio $2$ to $1$. How many integer values of k are there such that $0<k\leq 2007$ and the area between the parabola $y=k-x^2$ and the $x$-axis is an integer?

[asy] import graph; 	size(300); 	defaultpen(linewidth(0.8)+fontsize(10)); 	real k=1.5; 	real endp=sqrt(k); 	real f(real x) { 	return k-x^2; 	} 	path parabola=graph(f,-endp,endp)--cycle; 	filldraw(parabola, lightgray); 	draw((endp,0)--(endp,k)--(-endp,k)--(-endp,0)); 	label("Region I", (0,2*k/5)); 	label("Box II", (51/64*endp,13/16*k)); 	label("area(I) = $\frac23$\,area(II)",(5/3*endp,k/2));[/asy]

Solution

[asy] import graph; 	size(300); 	defaultpen(linewidth(0.8)+fontsize(10)); 	real k=1.5; 	real endp=sqrt(k); 	real f(real x) { 	return k-x^2; 	} 	path parabola=graph(f,-endp,endp)--cycle; 	filldraw(parabola, lightgray); 	draw((endp,0)--(endp,k)--(-endp,k)--(-endp,0)); 	label("Region I", (0,2*k/5)); 	label("Box II", (51/64*endp,13/16*k)); 	label("area(I) = $\frac23$\,area(II)",(0,k),N);  	label("$2 \sqrt{k}$",(0,0),S); 	label("$k$",(endp,k/2),E); [/asy]

The vertex of the quadratic is $(0,k)$. Factoring the quadratic results in $(\sqrt{k} + x)(\sqrt{k} - x)$, so the two zeroes are $(-\sqrt{k},0)$ and $(\sqrt{k},0)$. Thus, the area of the rectangle is $2k\sqrt{k}$, so the area of the parabola is $\frac{4}{3}k \sqrt{k}$.

In order for $\frac{4}{3}k \sqrt{k}$ to be an integer, $k$ must be a perfect square and a multiple of $3$. Note that $45^2 = 2025$, so the values that work are $3^2, 6^2, 9^2 \cdots 42^2$. In total, there are $\boxed{14}$ values of $k$ that satisfy the conditions.

See Also

2007 iTest (Problems)
Preceded by:
Problem 31
Followed by:
Problem 33
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