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

Latest revision as of 20:21, 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.

2007 iTest (Problems, Answer Key)
Preceded by: Problem 31	Followed by: Problem 33
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

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

Latest revision as of 20:21, 10 June 2018

Problem

Solution

See Also

@@ Line 19: / Line 19: @@
 	label("area(I) = $\frac23$\,area(II)",(5/3*endp,k/2));</asy>
-== Solution ==
+==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 <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>.
+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.
+==See Also==
+{{iTest box|year=2007|num-b=31|num-a=33}}
+[[Category:Introductory Algebra Problems]]
+[[Category:Introductory Number Theory Problems]]