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Geometry Difficulty 3.1 AMC 10/12 Prove it Canada

IMG0 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 $C$ .

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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. Thus, $|APXS| = wy$ , $|PDRX| = xy$ , $|SXBQ| = wz$ , and $|XRCQ| = xz$ . 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

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Figure for this problem $d = \dfrac{4t + \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} - \dfrac{4t - \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} = \sqrt{(-4t)^2 - 4(5t^2 - 6t)}$ From this we see that

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Figure for this problem d is as large as possible exactly when the discriminant is as large as possible. Here, the discriminant,

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Figure for this problem $\Delta$ $, is$ $\Delta = (-4t)^2 - 4(5t^2 - 6t) = 16t^2 - 20t^2 + 24t = -4t^2 + 24t$ $Completing the square,$ $\Delta = -4(t^2 - 6t) = -4(t^2 - 6t + 9 - 9) = -4(t^2 - 6t + 9) + 36 = -4(t-3)^2 + 36$ $Since$ (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

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Figure for this problem t = 3, which means that the distance between the

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Figure for this problem x $-intercepts is as large as possible when$ t = 3$.

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