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...") |
Rockmanex3 (talk | contribs) (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 == | + | ==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]] |
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 to . How many integer values of k are there such that and the area between the parabola and the -axis is an integer?
Solution
The vertex of the quadratic is . Factoring the quadratic results in , so the two zeroes are and . Thus, the area of the rectangle is , so the area of the parabola is .
In order for to be an integer, must be a perfect square and a multiple of . Note that , so the values that work are . In total, there are values of that satisfy the conditions.
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 31 |
Followed by: Problem 33 | |
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