2020 AMC 10B Problem 8

Problem

Points P and Q lie in a plane with PQ = 8. How many locations for point R in this plane are there such that the triangle with vertices P, Q, and R is a right triangle with area 12 square units?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

평면상에 PQ = 8 인 P , Q 점이 있다.
이 평면상에 꼭지점이 P, Q ,R 인 직각삼각형의 넓이가 12 인 꼭지점 R 은 몇개 있는가?

풀이)

삼각형의 넓이가 12니까 PQ 가 밑변이면 높이는 3 이다.

P점이 직각일때 R 이 2 점

Q점이 직각일때 R 이 2 점

R점이 직각일때 R 이 4 점
모두 8점이 있다.

(1) 가로 8 세로 3 직사각형 안에 2 대각선을 그리면
4개의 직각삼각형이 만들어 진다.
모두 넓이가 12 인 직각삼각형 이다.

(2) 반지름 4인 원을 그리면 지름이 8이다.
PQ 가 지름이 된다.
높이가 3인 원주상의 한점을 정하여
P점과 Q점을 연결하면 R점이 직각이 된다.
PQ의 위 아래 두점씩 4점의 직각이 있다.
4개의 직각삼각형이 만들어 진다.

넓이가 12 인, 8개의 직각삼각형을 만들수 있다.

답은 D (8) 이다.

Solution 1

There are 3 options here:

1. P is the right angle.

It's clear that there are 2 points that fit this, one that's directly to the right of P and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.

2. Q is the right angle.

Using the exact same reasoning, there are also 2 solutions for this one.

3. The new point is the right angle.

(Diagram temporarily removed due to asymptote error)

The diagram looks something like this. We know that the altitude to base $\overline{AB}$ must be $3$ since the area is $12$ . From here, we must see if there are valid triangles that satisfy the necessary requirements.

First of all, $\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24$ because of the area.

Next, $\overline{BC}^2+\overline{AC}^2=64$ from the Pythagorean Theorem.

From here, we must look to see if there are valid solutions. There are multiple ways to do this:

$\textbf{Recognizing min \& max:}$

We know that the minimum value of $\overline{BC}^2+\overline{AC}^2=64$ is when $\overline{BC} = \overline{AC} = \sqrt{24}$ . In this case, the equation becomes $24+24=48$ , which is LESS than $64$ . $\overline{BC}=1, \overline{AC} =24$ . The equation becomes $1+576=577$ , which is obviously greater than $64$ . We can conclude that there are values for $\overline{BC}$ and $\overline{AC}$ in between that satisfy the Pythagorean Theorem.

And since $\overline{BC} \neq \overline{AC}$ , the triangle is not isoceles, meaning we could reflect it over $\overline{AB}$ and/or the line perpendicular to $\overline{AB}$ for a total of $4$ triangles this case.

Solution 2

Note that line segment $\overline{PQ}$ can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for $Q$ that can satisfy the requirements - that being above or below $\overline{PQ}$ . As such, there are $2$ ways for this case. Similarly, one can find that there are also $2$ ways for point $Q$ to lie if $\overline{PQ}$ is the longer leg. If it is a hypotenuse, then there are $4$ possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}$ .

AOPS

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