AMC12 2020 A

AMC12 2020 A · Q24

AMC12 2020 A · Q24. It mainly tests Triangles (properties), Coordinate geometry.

Suppose that $\triangle ABC$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP = 1$, $BP = \sqrt{3}$, and $CP = 2$. What is $s$?

假设$\triangle ABC$是边长为$s$的等边三角形，具有唯一一点$P$在三角形内部使得$AP = 1$，$BP = \sqrt{3}$，$CP = 2$。求$s$？

(A) $1 + \sqrt{2}$ $1 + \sqrt{2}$

(B) $\sqrt{7}$ $\sqrt{7}$

(D) $\sqrt{5 + \sqrt{5}}$ $\sqrt{5 + \sqrt{5}}$

(E) $2\sqrt{2}$ $2\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Place the points in the coordinate plane so that $A=(0,0),\quad B=(s,0),\quad C=\left(\frac{s}{2},\frac{s}{2}\sqrt{3}\right),\quad \text{and}\quad P=(x,y).$ It is given that $AP^2=1=x^2+y^2,$ (1) $BP^2=3=(x-s)^2+y^2,$ and (2) $CP^2=4=\left(x-\frac{s}{2}\right)^2+\left(y-\frac{\sqrt{3}s}{2}\right)^2.$ (3) Subtracting the first equation from the other two equations gives $-2xs+s^2=2$ and (4) $-xs-\sqrt{3}ys+s^2=3.$ (5) From equation (4) it follows that $-xs=1-\frac{1}{2}s^2.$ (6) Substituting this into equation (5) leads to $y=\frac{s^2-4}{2\sqrt{3}s}.$ Equation (6) can be used to express $x$ in terms of $s$. Inserting these values for $x$ and $y$ into equation (1) produces an equation that $s$ satisfies $AP^2=1=x^2+y^2=\left(\frac{s^2-2}{2s}\right)^2+\left(\frac{s^2-4}{2\sqrt{3}s}\right)^2.$ This simplifies to $s^4-8s^2+7=0$, so $s^2=1$ or $7$. If $s=1$, then the point $P$ exists, but is outside the triangle. Hence $s=\sqrt{7}$, and indeed $P$ is inside the triangle.

答案（B）：在坐标平面中放置各点，使得 $A=(0,0),\quad B=(s,0),\quad C=\left(\frac{s}{2},\frac{s}{2}\sqrt{3}\right),\quad \text{且}\quad P=(x,y)。$ 已知 $AP^2=1=x^2+y^2,$（1） $BP^2=3=(x-s)^2+y^2,$（2） $CP^2=4=\left(x-\frac{s}{2}\right)^2+\left(y-\frac{\sqrt{3}s}{2}\right)^2.$（3）用后两式分别减去第一式得到 $-2xs+s^2=2$（4）以及 $-xs-\sqrt{3}ys+s^2=3.$（5）由（4）式可得 $-xs=1-\frac{1}{2}s^2.$（6）将其代入（5）式得到 $y=\frac{s^2-4}{2\sqrt{3}s}.$ （6）式可用来把$x$表示为$s$的函数。把$x$与$y$的这些表达式代入（1）式，得到$s$满足的方程 $AP^2=1=x^2+y^2=\left(\frac{s^2-2}{2s}\right)^2+\left(\frac{s^2-4}{2\sqrt{3}s}\right)^2.$ 化简得 $s^4-8s^2+7=0$，所以 $s^2=1$ 或 $7$。若 $s=1$，则点$P$存在，但在三角形外。因此 $s=\sqrt{7}$，并且此时$P$确在三角形内。

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