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Problem 516

AMC 10/12, early questions

Geometry Difficulty 3.1 Prove it CEMC Euclid · Canada · 2024

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 $m>0$ 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 $C$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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

[[IMAGE0]]

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| = $\text{2}$ 6}{3} =
4$.

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

If $|APXS| = 6$ , $|PDRX| = 2$ , and $|SWQB| = 3$ , then $a = |XRQC| = $\text{6}$ 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 = $\text{4t (-4t) 2}$ - 4(5t^2 - 6t)}}{2} The distance, $d$ , between these intercepts is their
difference, which is d = $\text{4t + (-4t) 2}$ - 4(5t^2 - 6t)}}{2} - $\text{4t - (-4t) 2}$ - 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 $t = 3$ .

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