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AMC8每日一题(2001年真题#14)

  • 2018-08-01     
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  今天课窝小编为大家整理了AMC8真题练习,希望考生们认真阅读,能够对你的考试有所帮助。


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2001 AMC 8 Problems/Problem 23

Problem

Points $R$$S$ and $T$ are vertices of an equilateral triangle, and points $X$$Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?

[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution

There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triples of $3$ of the $6$points. However, some of these created congruent triangles, and some don't even make triangles at all.

Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself.

Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$.

Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$.

Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$.

However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$. Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{\text{D}}$

问题

$ R $$ S $并且$ T $是等边三角形的顶点和点$ X $$ Y $并且$ Z $是其边的中点。使用这六个点中的任何三个作为顶点可以绘制多少个非一致三角形?

[asy]对SS,R,T,X,Y,Z;  SS =(2,2 * sqrt(3));  R =(0,0);  T =(4,0);  X =(2,0);  Y =(1,sqrt(3));  Z =(3,sqrt(3));  点(SS);  点(R);  点(T);  点(X);  点(Y);  点(Z);  标签( “$ S $”,SS,N);  标签( “$ R $”,R,SW);  标签( “$ T $”,T,SE);  标签( “$ X $”,X,S);  标签( “$ Y $”,Y,NW);  标签( “$ Z $”,Z,NE);  [/ ASY]

$ \ text {(A)} \ 1 \ qquad \ text {(B)} \ 2 \ qquad \ text {(C)} \ 3 \ qquad \ text {(D)} \ 4 \ qquad \ text {(E )} \ 20 $

$ 6 $图中的点,$ 3 $他们都需要形成一个三角形,所以有$ {6 \选择{3}} = 20 $可能的三元组$ 3 $的的$ 6 $点。然而,其中一些创建了全等三角形,有些甚至根本不制作三角形。

案例1:三角形一致,$ \ triangle RST $显然只有$ 1 $这些:$ \ triangle RST $本身。

案例2:三角形一致,$ \ triangle SYZ $有以下$ 4 $几种:$ \三角形SYZ,\三角形RXY,\三角形TXZ,$$ \ triangle XYZ $

案例3:三角形一致,$ \ triangle RSX $有以下$ 6 $几种:$ \ triangle RSX,\ triangle TSX,\ triangle STY,\ triangle RTY,\ triangle RSZ,$$ \ triangle RTZ $

案例4:三角形一致,$ \ triangle SYX $还有$ 6 $这些:$ \ triangle SYX,\ triangle SZX,\ triangle TYZ,\ triangle TYX,\ triangle RXZ,$$ \ triangle RYZ $

但是,如果我们添加这些,我们只$ 1 + 4 + 6 + 6 = 17 $考虑$ 20 $可能的三元组。我们看到剩下的三胞胎甚至不形成三角形; 他们是$ SYR,RXT,$$ $先令。将这些添加$ 3 $到所有可能的三元组的总产量中,因此我们看到只有$ 4 $可能的非一致,非简并三角形,$ \盒装{\ {文字d}} $

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