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Calculus Level 5

Consider the parabola $P$ with the equation $x^{2} - 8x + 4y = 0.$ A line $y = b$ , with $0 \lt b \lt 4$ , is then drawn creating two regions $R_{1}$ and $R_{2}$ of equal area $A$ where

$R_{1}$ is the region bounded below by the line $y = b$ and above by the curve $P$ , and
$R_{2}$ is the region in the first quadrant bounded above by the line $y = b$ , to the left by the $y$ -axis and to the right by the branch of $P$ nearest the $y$ -axis.

If $A = \dfrac{a}{c}$ , where $a$ and $c$ are positive coprime integers, then find $a + b + c.$

The answer is 14.

1 solution

Brian Charlesworth
Feb 16, 2015

The equation for the parabola can be written as $y = -\frac{1}{4}(x - 4)^{2} + 4.$

With $(k,b)$ being the coordinates of the leftmost point of intersection of $P$ and the line $y = b$ , we have that $b = -\frac{1}{4}(k - 4)^{2} + 4.$

Then to have the areas of regions $R_{1}$ and $R_{2}$ be equal, we require that

$\displaystyle\int_{0}^{k} (b + \frac{1}{4}(x - 4)^{2} + 4) dx = 2\int_{k}^{4} (-\frac{1}{4}(x - 4)^{2} + 4 - b) dx$ ,

where I made use of the symmetry of $P$ about the line $x = 4$ . Performing the integration and simplifying, we have that

$(b - 4)k + \frac{1}{12}(k - 4)^{3} + \frac{16}{3} = 2[(4 - b)(4 - k) + \frac{1}{12}(k - 4)^{3}],$

which, after substituting $b - 4 = -\frac{1}{4}(k - 4)^{2}$ , yields

$-\frac{k}{4}(k - 4)^{2} + \frac{5}{12}(k - 4)^{3} + \frac{16}{3} = 0$

$\Longrightarrow -3k(k - 4)^{2} + 5(k - 4)^{3} + 64 = 0$

$\Longrightarrow k^{3} - 18k^{2} + 96k - 128 = 0 \Longrightarrow (k - 2)(k - 8)^{2} = 0.$

Now we are looking for $k \lt 4$ , i.e., to the left of the vertex of $P$ , so we take $k = 2$ , which in turn gives us $b = 3$ . Plugging these values into one of the above area integrals yields the value $A = \frac{8}{3}.$

Thus $a + b + c = 8 + 3 + 3 = \boxed{14}.$

Upstairs, downstairs .....

The answer is 14.

1 solution

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