AMC12 2004 B

AMC12 2004 B · Q14

AMC12 2004 B · Q14. It mainly tests Triangles (properties), Similarity.

In $\triangle ABC$, $AB=13$, $AC=5$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=4$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?

在$\triangle ABC$中，$AB=13$，$AC=5$，$BC=12$。点$M$和$N$分别在$AC$和$BC$上，且$CM=CN=4$。点$J$和$K$在$AB$上，使得$MJ$和$NK$都垂直于$AB$。五边形$CMJKN$的面积是多少？

(A) 15 15

(B) \frac{81}{5} \frac{81}{5}

(D) \frac{240}{13} \frac{240}{13}

(E) 20 20

Answer

Correct choice: (D)

正确答案：（D）

Solution

The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon. Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below: Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$. We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$. Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$. Now for the smaller triangles: We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$. Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$. Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$. Observe that all of the triangles in the problem are right triangles and similar with a ratio of 5-12-13. The largest triangle $\Delta ABC$ has area $\dfrac{5\cdot12}2=30$. $\Delta AMJ$ is linearly scaled down from $\Delta ABC$ by a factor of $\dfrac{5-4}{13}=\dfrac1{13}$ (as we can see from comparing the two hypotenuses), while $\Delta NBK$ is scaled by a factor of $\dfrac{12-4}{13}=\dfrac8{13}$. The area we desire is the combined area of $\Delta AMJ$ and $\Delta NBK$ subtracted from the area of $\Delta ABC$, which is \[30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.\] Split the pentagon along a different diagonal as follows:

三角形$ABC$显然是直角三角形，其面积为$\frac{5\cdot 12}2 = 30$。若能求出三角形$AMJ$与$BNK$的面积，就可以用总面积减去它们得到五边形的面积。作从$C$到$AB$的高$CL$。由于$AB=13$且面积为$30$，得$CL=\frac{60}{13}$。如下图所示：注意到三角形$ABC$、$AMJ$、$ACL$、$CBL$与$NBK$的角都相同，因此它们相似。我们已知其中一些边长，将利用这些信息计算面积。若两个图形相似比为$k$，则面积比为$k^2$。我们将反复使用这一事实。下面用$[XYZ]$表示三角形$XYZ$的面积。有$\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$，因此$[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$。又$\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$，因此$[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$。再看较小的三角形：已知$\frac{AM}{AC}=\frac 15$，因此$[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$。同理，$\frac{BN}{BC}=\frac 8{12} = \frac 23$，因此$[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$。最后，五边形面积为$30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$。注意：题中所有三角形都是直角三角形，且都按$5$-$12$-$13$的比例相似。最大三角形$\Delta ABC$面积为$\dfrac{5\cdot12}2=30$。$\Delta AMJ$相对于$\Delta ABC$的线性缩放因子为$\dfrac{5-4}{13}=\dfrac1{13}$（可由两条斜边比较看出），而$\Delta NBK$的缩放因子为$\dfrac{12-4}{13}=\dfrac8{13}$。所求面积为从$\Delta ABC$面积中减去$\Delta AMJ$与$\Delta NBK$的面积之和，即 \[30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.\] 沿另一条对角线将五边形分割如下：

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