2014 AMC 10A Problem 14

P11-15~4 minPrint

Coordinate Geometry Circles Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$ ?

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Key insight

Find the two slopes: one is m, the other is -1/m. Making the two y-intercepts add to zero gives m = 3, so the intercepts are 10 and -10.

Solution

The plan: find the two lines, read off their $y$ -intercepts, then find the area.

1. Name the slopes. Call the slope of one line $m$ . Perpendicular slopes multiply to $-1$ , so the other line has slope $-\frac1m$ .
2. Find the intercepts. A line through $(6, 8)$ with slope $m$ is $y = 8 + m(x - 6)$ . Put $x = 0$ : its $y$ -intercept is $8 - 6m$ . The same way, the other line's $y$ -intercept is $8 + \frac{6}{m}$ .
3. Use "the sum is zero".
$(8 - 6m) + \left(8 + \frac{6}{m}\right) = 0 \quad\Longrightarrow\quad 16 = 6m - \frac{6}{m}.$
Multiply by $m$ : $16m = 6m^2 - 6$ . Divide by $2$ and rearrange: $3m^2 - 8m - 3 = 0$ .
4. Solve. This factors as $(3m + 1)(m - 3) = 0$ , so $m = 3$ or $m = -\frac13$ . These are just the two lines swapped, so take $m = 3$ .
5. The intercepts. $8 - 6 \cdot 3 = -10$ and $8 + \frac63 = 10$ . So $P$ and $Q$ are $(0, 10)$ and $(0, -10)$ . Their sum is $0$ , as required.
6. The area. Use $PQ$ as the base; it lies on the $y$ -axis and has length $20$ . The height is how far $A$ is from the $y$ -axis, which is its $x$ -coordinate, $6$ .
$\text{area} = \frac12 \cdot 20 \cdot 6 = 60.$

The answer is $\boxed{\textbf{(D)}\ 60}$ .

Why this works

"Perpendicular" becomes "slopes $m$ and $-\frac1m$ ", and "intercepts add to zero" becomes one equation in $m$ . After that the problem is ordinary algebra. For the area, choose the side that lies on an axis as the base, because then the height is simply a coordinate of the third point.

Alternative approach

A faster way uses a circle. The intercepts add to zero, so $P = (0, p)$ and $Q = (0, -p)$ , and the origin $O$ is exactly halfway between them. The angle at $A$ is a right angle. A known fact says: if a right angle stands on a segment, its corner lies on the circle that has that segment as a diameter. That circle has center $O$ and radius $p$ , so $A$ is at distance $p$ from $O$ :
$p = \sqrt{6^2 + 8^2} = 10.$
The base $PQ$ is $20$ , the height is $6$ , and the area is $60$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write $P = (0, p)$ and $Q = (0, -p)$ with $p > 0$ ; the origin $O$ is the midpoint of $\overline{PQ}$ .

Since the lines $AP$ and $AQ$ are perpendicular, $\angle PAQ = 90^\circ$ , so $A$ lies on the circle whose diameter is $\overline{PQ}$ . That circle is centered at $O$ with radius $p$ , hence
$p = OA = \sqrt{6^2 + 8^2} = 10.$

Now $\triangle APQ$ has base $PQ = 20$ along the $y$ -axis, and its height is the horizontal distance from $A$ to the $y$ -axis, which is $6$ . Therefore
$[APQ] = \frac12 \cdot 20 \cdot 6 = 60.$

The answer is $\boxed{\textbf{(D)}\ 60}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)