Difference between revisions of "2007 iTest Problems/Problem 32"

(Created page with "== 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 region...")
 
(Solution to Problem 32)
 
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label("area(I) = $\frac23$\,area(II)",(5/3*endp,k/2));</asy>
 
label("area(I) = $\frac23$\,area(II)",(5/3*endp,k/2));</asy>
  
== Solution ==
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==Solution==
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<asy>
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import graph;
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size(300);
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defaultpen(linewidth(0.8)+fontsize(10));
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real k=1.5;
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real endp=sqrt(k);
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real f(real x) {
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return k-x^2;
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}
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path parabola=graph(f,-endp,endp)--cycle;
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filldraw(parabola, lightgray);
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draw((endp,0)--(endp,k)--(-endp,k)--(-endp,0));
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label("Region I", (0,2*k/5));
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label("Box II", (51/64*endp,13/16*k));
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label("area(I) = $\frac23$\,area(II)",(0,k),N);
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label("$2 \sqrt{k}$",(0,0),S);
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label("$k$",(endp,k/2),E);
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</asy>
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The vertex of the quadratic is <math>(0,k)</math>.  Factoring the quadratic results in <math>(\sqrt{k} + x)(\sqrt{k} - x)</math>, so the two zeroes are <math>(-\sqrt{k},0)</math> and <math>(\sqrt{k},0)</math>.  Thus, the area of the rectangle is <math>2k\sqrt{k}</math>, so the area of the parabola is <math>\frac{4}{3}k \sqrt{k}</math>.
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In order for <math>\frac{4}{3}k \sqrt{k}</math> to be an integer, <math>k</math> must be a perfect square and a multiple of <math>3</math>.  Note that <math>45^2 = 2025</math>, so the values that work are <math>3^2, 6^2, 9^2 \cdots 42^2</math>.  In total, there are <math>\boxed{14}</math> values of <math>k</math> that satisfy the conditions.
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==See Also==
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{{iTest box|year=2007|num-b=31|num-a=33}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 21:21, 10 June 2018

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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