AMC12 2015 B

AMC12 2015 B · Q24

AMC12 2015 B · Q24. It mainly tests Circle theorems, Coordinate geometry.

Four circles, no two of which are congruent, have centers at $A, B, C,$ and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\frac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\frac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $PQ$. What is $AR + BR + CR + DR$?

四个圆，不两两全等的，圆心在$A, B, C,$和$D$，点$P$和$Q$在所有四个圆上。圆$A$的半径是圆$B$半径的$\frac{5}{8}$，圆$C$的半径是圆$D$半径的$\frac{5}{8}$。此外，$AB = CD = 39$且$PQ = 48$。让$R$为$PQ$的中点。$AR + BR + CR + DR$是多少？

(A) 180 180

(B) 184 184

(D) 192 192

(E) 196 196

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Points $A$, $B$, $C$, $D$, and $R$ all lie on the perpendicular bisector of $\overline{PQ}$. Assume $R$ lies between $A$ and $B$. Let $y=AR$ and $x=\frac{AP}{5}$. Then $BR=39-y$ and $BP=8x$, so $y^2+24^2=25x^2$ and $(39-y)^2+24^2=64x^2$. Subtracting the two equations gives $x^2=39-2y$, from which $y^2+50y-399=0$, and the only positive solution is $y=7$. Thus $AR=7$, and $BR=32$. Note that circles $A$ and $B$ are determined by the assumption that $R$ lies between $A$ and $B$. Thus because the four circles are noncongruent, $R$ does not lie between $C$ and $D$. Let $w=CR$ and $z=\frac{CP}{5}$. Then $DR=39+w$ and $DP=8z$, so $w^2+24^2=25z^2$ and $(39+w)^2+24^2=64z^2$. Subtracting the two equations gives $z^2=39+2w$, from which $w^2-50w-399=0$, and the only positive solution is $w=57$. Thus $CR=57$ and $DR=96$. Again, the uniqueness of the solution implies that $R$ must indeed lie between $A$ and $B$. The requested sum is $7+32+57+96=192$.

答案（D）：点 $A$、$B$、$C$、$D$ 和 $R$ 都在 $\overline{PQ}$ 的垂直平分线上。假设 $R$ 在 $A$ 与 $B$ 之间。令 $y=AR$，$x=\frac{AP}{5}$。则 $BR=39-y$ 且 $BP=8x$，所以 $y^2+24^2=25x^2$ 以及 $(39-y)^2+24^2=64x^2$。两式相减得 $x^2=39-2y$，从而 $y^2+50y-399=0$，唯一的正解是 $y=7$。因此 $AR=7$，$BR=32$。注意：圆 $A$ 和圆 $B$ 是由“$R$ 在 $A$ 与 $B$ 之间”这一假设确定的。因此由于四个圆不全等，$R$ 不在 $C$ 与 $D$ 之间。令 $w=CR$，$z=\frac{CP}{5}$。则 $DR=39+w$ 且 $DP=8z$，所以 $w^2+24^2=25z^2$ 以及 $(39+w)^2+24^2=64z^2$。两式相减得 $z^2=39+2w$，从而 $w^2-50w-399=0$，唯一的正解是 $w=57$。因此 $CR=57$，$DR=96$。同样，由解的唯一性可知 $R$ 的确必须在 $A$ 与 $B$ 之间。所求的和为 $7+32+57+96=192$。

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