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考题22 2015 AMC 10B

2018-08-06 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：

Problem 22

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$ . What is $FG + JH + CD$ ?

$考题22 2015 AMC 10B$

Solution 1

Triangle $AFG$ is isosceles, so $AG=AF=1$ . $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$ , notice that $\triangle JHG \cong \triangle AFG$ . Therefore, $JH=AF=1$ .

Since $\triangle AJH \sim \triangle AFG$ ,

$考题22 2015 AMC 10B$

$考题22 2015 AMC 10B$ $1 = FG^2 + FG$ $FG^2+FG-1 = 0$ $FG = \frac{-1 \pm \sqrt{5} }{2}$

So, $FG=\frac{-1 + \sqrt{5}}{2}$ since $FG$ must be greater than 0.

Notice that $考题22 2015 AMC 10B$ .

Therefore, $FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$

Solution 2 (Trigonometry)

Note that since $ABCDE$ is a regular pentagon, all of its interior angles are $108^\circ$ . We can say that pentagon $FGHIJ$ is also regular by symmetry. So, all of the interior angles of $FGHIJ$ are $108^\circ$ . Now, we can angle chase and use trigonometry to get that $FG=2\sin18^\circ$ , $JH=2\sin18^\circ*(2\sin18^\circ+1)$ , and $DC=2\sin18^\circ*(2\sin18^\circ+2)$ . Adding these together, we get that $FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)$ . Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to $8\sin18^\circ*(1+\sin18^\circ)$ , but we can find that this is closest to $考题22 2015 AMC 10B$ .

Solution 3

When you first see this problem you can't help but see similar triangles. But this shape is filled with $36 - 72 - 72$ triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of $FG$ so we can apply similar triangles easily. To simplify the process lets write $FG$ as $x$ .

First what is $JH$ in terms of $x$ , also remember $AJ = 1+x$ :

$\frac{JH}{1+x}=\frac{x}{1}$

$JH$ = ${x}^2+x$

Next, find $DC$ in terms of $x$ , also remember $AD = 2+x$ :

$\frac{DC}{2+x}=\frac{x}{1}$

$DC$ = ${x}^2+2x$

So adding all the $FG + JH + CD$ we get $2{x}^2+4x$ . Now we have to find out what x is. For this, we break out a bit of trig. Let's look at $\triangle AFG$ By the law of sines: