AMC 8 2024 Problem 11
AMC8 2024 年第 11 题
坐标几何★★★☆☆
The coordinates of △ABC are A(5, 7), B(11, 7), and C(3, y), with y > 7.
The area of △ABC is 12.
What is the value of y?
△ABC 的顶点坐标分别为 A(5,7)、B(11,7)和 C(3,y)，其中 y>7。已知△ABC 的面积为12，请问y的值为多少？
- A.8
- B.9
- C.10
- D.11
- E.12
正确答案 · Correct Answer
D
解析 · Solution
解法1
由于三角形的底边长为6，我们可以将该值代入作为底边。然后，我们求解方程得出高。计算过程如下：这意味着h=4，因此我们需要在坐标上加4。所以答案是7+4=11。
鞋带定理， 的面积为
根据题意，这个值等于 12。现在我们求解 y。 ，|6y-42| = 24 ，6y - 42 = 24 或 6y - 42 = -24 ，6y = 66 或 6y = 18 ，y = 11 或 y = 3 。 然而，由于题目中说明 y > 7，我们唯一的有效解是 11。
如图所示，该三角形由向量
$\begin{bmatrix} \\ y-7\end{bmatrix}
$ 和 $\begin{bmatrix}6\0\end{bmatrix}
$ 确定。回忆一下，这些向量的行列式的绝对值是由它们确定的平行四边形的面积；三角形的面积是该平行四边形面积的一半。因此，必须有 $$ 。展开行列式，我们得到 $-6(y-7) = 24$ 或 $-6(y-7) = -24$ 。分别解每个方程，得到 $y = 3$ 或 $y = 11$ 。然而，题目指出 $y > 7$ ，因此唯一有效的解是 11 。（再次！）
画一个矩形，使斜边是矩形的对角线。底边为8，高为 y - 7 ，所以面积为。在矩形内部，除了原三角形外，还有两个额外的直角三角形。较大的直角三角形的面积是矩形面积的一半，因此它的面积为。较小的直角三角形的底边为2，高为 y - 7 ，所以它的面积为。从矩形面积中减去这两个额外的直角三角形面积，得到。由于题目告诉我们原三角形的面积为12，因此得到方程 ， ， 。所以问题的答案是D。
Solution 1
Since the triangle has a base of 6, we can plug in that value as the base. Then, we can solve the equation for the height. Doing so gives us, This means that h=4, so that means that we have to add 4 to the So the answer is 7+4=11
Shoelace Theorem, has area
From the problem, this is equal to 12. We now solve for y. , |6y-42| = 24, 6y - 42 = 246y - 42 = -24, 6y = 666y = 18, y = 11y = 3. However, since, as stated in the problem, y > 7, our only valid solution is 11.
As in the figure, the triangle is determined vectors
$\begin{bmatrix} \\ y-7\end{bmatrix}
$ and $\begin{bmatrix}6\0\end{bmatrix}
$. Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined vectors; the triangle has half the area of that parallelogram. Then we must have that $$. Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$. Solving each equation individually, we find that $y = 3$ or $y = 11$. However, the problem states that $y > 7$, so the only valid solution is 11. (again!)
Draw a rectangle so that the hypotenuse is the diagonal of the rectangle. The base is 8 and the height is y - 7, so the area is Inside the rectangle, there are 2 extra right triangles along with the original triangle. The larger right triangle has an area of half the rectangle, so it has an area of The smaller right triangle has a base of 2 and a height of y - 7, so its area is Subtracting the extra right triangles from the area of the rectangle, you get Since the problem told us that the original triangle had an area of 12, you get the equation , , . So the answer to the problem is D.
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