AMC12 2021 B

AMC12 2021 B · Q24

AMC12 2021 B · Q24. It mainly tests Quadratic equations, Triangles (properties).

Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points. Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$

设 $ABCD$ 是面积为 $15$ 的平行四边形。点 $P$ 和 $Q$ 分别是 $A$ 和 $C$ 在直线 $BD$ 上的投影；点 $R$ 和 $S$ 分别是 $B$ 和 $D$ 在直线 $AC$ 上的投影。见图，该图还显示了这些点的相对位置。假设 $PQ=6$ 和 $RS=8$，设 $d$ 表示 $\overline{BD}$ 的长度，即 $ABCD$ 的较长对角线。然后 $d^2$ 可写成 $m+n\sqrt p$ 的形式，其中 $m,n,p$ 是正整数，且 $p$ 没有被任何质数的平方整除。求 $m+n+p$？

(A) 81 81

(B) 89 89

(D) 105 105

(E) 113 113

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Now note that since $\angle APB = \angle ARB = 90^\circ$, quadrilateral $ARPB$ is cyclic, and so \[XR\cdot XA = XP\cdot XB,\]which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$. Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15\]Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$, and so\[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\]The requested answer is $32 + 8 + 41 = \boxed{\textbf{(A)} ~81}$.

设 $X$ 为对角线 $AC$ 和 $BD$ 的交点。注意到由对称性 $X$ 是 $\overline{PQ}$ 和 $\overline{RS}$ 的中点，因此 $XP = XQ = 3$ 和 $XR = XS = 4$。现在注意到由于 $\angle APB = \angle ARB = 90^\circ$，四边形 $ARPB$ 是圆内接的，因此 \[XR\cdot XA = XP\cdot XB,\] 这意味着 $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$。因此设 $x> 0$ 使得 $XA = 3x$ 和 $XB = 4x$。在 $\triangle APX$ 上用勾股定理得 $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$，因此 \[ [ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15 \] 解此关于 $x^2$ 的方程得 $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$，因此 \[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\] 所求答案是 $32 + 8 + 41 = \boxed{\textbf{(A)} ~81}$。

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