AMC12 2024 B

AMC12 2024 B · Q20

AMC12 2024 B · Q20. It mainly tests Vieta / quadratic relationships (basic), Area & perimeter.

Suppose $A$, $B$, and $C$ are points in the plane with $AB=40$ and $AC=42$, and let $x$ be the length of the line segment from $A$ to the midpoint of $\overline{BC}$. Define a function $f$ by letting $f(x)$ be the area of $\triangle ABC$. Then the domain of $f$ is an open interval $(p,q)$, and the maximum value $r$ of $f(x)$ occurs at $x=s$. What is $p+q+r+s$?

假设 $A$、$B$ 和 $C$ 是平面上的点，$AB=40$，$AC=42$，设 $x$ 是从 $A$ 到 $\overline{BC}$ 中点的线段长度。定义函数 $f$，使得 $f(x)$ 是 $\triangle ABC$ 的面积。那么 $f$ 的定义域是一个开区间 $(p,q)$，最大值 $r$ 在 $x=s$ 处取得。求 $p+q+r+s$？

(A) 909 909

(B) 910 910

(D) 912 912

(E) 913\qquad 913\qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let the midpoint of $BC$ be $M$, and let the length $BM = CM = a$. We know there are limits to the value of $x$, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length $BC$ to $AC$ and $AB$, and doesn't contain any information about the median. Therefore we're going to have to write the side $BC$ in terms of $x$ and then use the triangle inequality to find bounds on $x$. We use Stewart's theorem to relate $BC$ to the median $AM$: $man + dad = bmb + cnc$. In this case $m = \frac{a}2$, $n=\frac{a}2$, $a = m+n$, $d = x$, $b = 42$, $c = 40$. Therefore we get the equation $2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2$ $2a^2 + 2x^2 = 42^2 + 40^2$. Notice that since $20-21-29$ is a pythagorean triple, this means $2a^2 + 2x^2 = 58^2$. \[\implies a^2 = \frac{58^2}{2}-x^2\] \[\implies a = \sqrt{\frac{58^2}{2}-x^2}\] By triangle inequality, $2a+40>42 \implies a>1$ and $40+42>2a \implies a<41$ Let's tackle the first inequality: \[\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1\] \[\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2\] Here we use the property that $\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2$. Therefore in this case, $x<41$. For the second inequality, \[\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2\] \[\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2\] \[\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1\] Therefore we have $1<x<41$, so the domain of $f(x)$ is $(1,41)$. The area of this triangle is $\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)$. The maximum value of the area occurs when the triangle is right, i.e. $\theta = 90^{\circ}$. Then the area is $\frac{1}{2} \cdot 40 \cdot 42 = 840$. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is $40^2+42^2 = 58^2$. Thus the length of $x$ is $29$. Our final answer is $1+41+840+29 = \boxed{\textbf{911 } (C)}$

设 $BC$ 的中点为 $M$，$BM = CM = a$。我们知道 $x$ 的值有界限，这些界限可能通过三角不等式找到。但三角不等式涉及第三边长 $BC$ 与 $AC$ 和 $AB$ 的关系，不包含关于中线的任何信息。因此我们需要用 $x$ 表示边 $BC$，然后用三角不等式找到 $x$ 的界限。使用 Stewart 定理将 $BC$ 与中线 $AM$ 关联：$man + dad = bmb + cnc$。在本例中 $m = \frac{a}2$，$n=\frac{a}2$，$a = m+n$，$d = x$，$b = 42$，$c = 40$。因此得到方程 $2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2$ $2a^2 + 2x^2 = 42^2 + 40^2$。注意到 $20-21-29$ 是勾股三元组，这意味着 $2a^2 + 2x^2 = 58^2$。 \[\implies a^2 = \frac{58^2}{2}-x^2\] \[\implies a = \sqrt{\frac{58^2}{2}-x^2}\] 由三角不等式，$2a+40>42 \implies a>1$ 且 $40+42>2a \implies a<41$ 处理第一个不等式：\[\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1\] \[\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2\] 这里使用性质 $\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2$。因此在本例中，$x<41$。对于第二个不等式，\[\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2\] \[\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2\] \[\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1\] 因此我们有 $1<x<41$，所以 $f(x)$ 的定义域是 $(1,41)$。该三角形的面积是 $\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)$。面积的最大值发生在直角三角形时，即 $\theta = 90^{\circ}$。此时面积为 $\frac{1}{2} \cdot 40 \cdot 42 = 840$。直角三角形的中线长度为其斜边的一半，其平方为 $40^2+42^2 = 58^2$。因此 $x$ 的长度为 $29$。最终答案是 $1+41+840+29 = \boxed{\textbf{911 } (C)}$

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