Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 18"

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==Problem==
 
==Problem==
<div style="float:right">
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[[Image:2006 CyMO-18.PNG|250px|right]]
[[Image:2006 CyMO-18.PNG|250px]]
 
</div>
 
  
 
<math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>\Gamma (0,k)</math>.
 
<math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>\Gamma (0,k)</math>.
 
If the area if the rectangle <math>OAB\Gamma</math> is <math>8</math>, then the equation of the parabola is
 
If the area if the rectangle <math>OAB\Gamma</math> is <math>8</math>, then the equation of the parabola is
  
A. <math>y=\frac{1}{2}(x+2)^2</math>
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<math>\mathrm{(A)}\ y=\frac{1}{2}(x+2)^2\qquad\mathrm{(B)}\ y=\frac{1}{2}(x-2)^2\qquad\mathrm{(C)}\ y=x^2+2\qquad\mathrm{(D)}\ y=x^2-2x+1\qquad\mathrm{(E)}\ y=x^2-4x+4</math>
  
B. <math>y=\frac{1}{2}(x-2)^2</math>
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==Solution==
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Since the parabola is symmetric about the line <math>x = k</math>, <math>B</math> has coordinates <math>(2k,k)</math>. The area of the rectangle is <math>k \cdot 2k = 8 \Longrightarrow k = 2</math>, so the vertex is at <math>(2,0)</math>.
  
C. <math>y=x^2+2</math>
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Thus, the equation of the parabola is <math>y = a(x-2)^2</math>. Plugging in point <math>(0,2)</math>, we find <math>a = \frac{1}{2}</math>, and the answer is <math>\mathrm{B}</math>.
 
 
D. <math>y=x^2-2x+1</math>
 
 
 
E. <math>y=x^2-4x+4</math>
 
 
 
==Solution==
 
Since the parabola is symmetric about the line <math>x = k</math>, <math>B</math> has coordinates <math>(2k,k)</math>. The area of the rectangle is <math>k \cdot 2k = 8 \Longrightarrow k = 2</math>, so the vertex is at <math>(2,0)</math>. Thus the equation of the parabola is <math>y = a(x-2)^2</math>, and plugging in point <math>(0,2)</math> and finding <math>a = \frac{1}{2}</math>, the answer is <math>\mathrm{B}</math>.
 
  
 
==See also==
 
==See also==

Latest revision as of 12:39, 26 April 2008

Problem

2006 CyMO-18.PNG

$K(k,0)$ is the minimum point of the parabola and the parabola intersects the y-axis at the point $\Gamma (0,k)$. If the area if the rectangle $OAB\Gamma$ is $8$, then the equation of the parabola is

$\mathrm{(A)}\ y=\frac{1}{2}(x+2)^2\qquad\mathrm{(B)}\ y=\frac{1}{2}(x-2)^2\qquad\mathrm{(C)}\ y=x^2+2\qquad\mathrm{(D)}\ y=x^2-2x+1\qquad\mathrm{(E)}\ y=x^2-4x+4$

Solution

Since the parabola is symmetric about the line $x = k$, $B$ has coordinates $(2k,k)$. The area of the rectangle is $k \cdot 2k = 8 \Longrightarrow k = 2$, so the vertex is at $(2,0)$.

Thus, the equation of the parabola is $y = a(x-2)^2$. Plugging in point $(0,2)$, we find $a = \frac{1}{2}$, and the answer is $\mathrm{B}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 17
Followed by
Problem 19
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