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考题22 2015 AMC 10B

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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 22In the figure shown below,  is a regular pentagon and . What is ?

Solution 1Triangle  is isosceles, so .  since  is also isosceles. Using the symmetry of pentagon , notice that . Therefore, .

Since ,

So,  since  must be greater than 0.

Notice that .

Therefore,

Solution 2 (Trigonometry)Note that since  is a regular pentagon, all of its interior angles are . We can say that pentagon  is also regular by symmetry. So, all of the interior angles of  are . Now, we can angle chase and use trigonometry to get that , , and . Adding these together, we get that . 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 , but we can find that this is closest to .

Solution 3When you first see this problem you can´t help but see similar triangles. But this shape is filled with  triangles throwing us off. First, let us write our answer in terms of one side length. I chose to write it in terms of  so we can apply similar triangles easily. To simplify the process lets write  as .

First what is  in terms of , also remember :

=

Next, find  in terms of , also remember :

=

So adding all the  we get . Now we have to find out what x is. For this, we break out a bit of trig. Let´s look at  By the law of sines:

Now by the double angle identities in trig.  substituting in

A good thing to memorize for AMC and AIME is the exact values for all the nice sines and cosines. You would then know that: =

so now we know:

Substituting back into  we get

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