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Geometry Difficulty 3.1 AMC 10/12 Prove it Canada

In the diagram, $AB$ is perpendicular to $CD$ (with $B$ on $CD$ ), $CP$ is perpendicular to $AD$ (with $P$ on $AD$ ), and $N$ is the point of intersection of $AB$ and $CP$ .

Also, $\angle ADB = 45\degree$ ,
$AB=12$ , and $CB=6$ . What is the area of $\triangle APN$ ?
In the diagram, the line with equation
$y=-3x+6$ crosses the $x$ -axis at $A$ and the $y$ -axis at $B$ . Suppose that $\text{m gt;0}$ and that the line with equation
$y = mx + 1$ crosses the $y$ -axis at $D$ and intersects the line with equation
$y = -3x+6$ at the point $C$ .

If $O$ is the origin and the area
of $\triangle ACD$ is $\frac{1}{2}$ of the area of $\triangle ABO$ , determine the coordinates
of $C$ .

Figure for this problem

Solution

Suppose that $AP = w$ , $PD = x$ , $AS = y$ , and $SB = z$ .

We use the notation $|APXS|$ to
represent the area of $APXS$ , and so
on.

Then, $|APXS| \cdot |XRCQ| = wy\cdot xz = xy \cdot wz = |PDRX| \cdot |SXQB|$ If $|APXS| = 2$ , $|PDRX| = 3$ , and $|SWQB| = 6$ , then $a = |XRQC| = \dfrac{2 \cdot 6}{3} = 4$ .

If $|APXS| = 2$ , $|PDRX| = 6$ , and $|SWQB| = 3$ , then $a = |XRQC| = \dfrac{2 \cdot 3}{6} = 1$ .

If $|APXS| = 6$ , $|PDRX| = 2$ , and $|SWQB| = 3$ , then $a = |XRQC| = \dfrac{6 \cdot 3}{2} = 9$ .

Since we are told that there are three possible values for $a$ , then these are 1, 4 and 9.

(Can you explain why there are exactly three such values?)
The $x$ -intercepts of the
parabola with equation $y = x^2 - 4tx + 5t^2 - 6t$ are $x = \dfrac{4t \pm \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2}$ The distance, $d$ , between these intercepts is their
difference, which is $d = \dfrac{4t + \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} - \dfrac{4t - \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} = \sqrt{(-4t)^2 - 4(5t^2 - 6t)}$ From this we see that
$d$ is as large as possible exactly
when the discriminant is as large as possible. Here, the discriminant,
$\Delta$ , is $\Delta = (-4t)^2 - 4(5t^2 - 6t) = 16t^2 - 20t^2 + 24t = -4t^2 + 24t$ Completing the square, $\Delta = -4(t^2 - 6t) = -4(t^2 - 6t + 9 - 9) = -4(t^2 - 6t + 9) + 36 = -4(t-3)^2 + 36$ Since $(t-3)^2 \geq 0$ , then $\Delta \leq 36$ and $\Delta = 36$ exactly when $(t-3)^2 = 0$ or $t = 3$ .

Therefore, the discriminant is maximized when $t = 3$ , which means that the distance
between the $x$ -intercepts is as
large as possible when $t = 3$ .

Figure for this problem

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.