首页> 重点归纳 > AMC数学竞赛真题2016年12A 11-12

AMC数学竞赛真题2016年12A 11-12

2018-09-01 重点归纳

AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 11

Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$

Solution

Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.

Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.

From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$ .

Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$ .

Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$ .

Subtracting these equations, we get $a + b + c = 36$ .

Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}$

Solution 2

An easier way to solve the problem: Since $42$ students cannot sing, there are $100-42=58$ students who can.

Similarly $65$ students cannot dance, there are $100-65=35$ students who can.

And $29$ students cannot act, there are $100-29=71$ students who can.

Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories. There are no students who have all $3$ talents, nor those who have none $(0)$ , so only $1$ or $2$ talents are viable.

Thus, there are $164-100=\boxed{\textbf{(E) }64}$ students who have $2$ of $3$ talents.

Problem 12

In $\triangle ABC$ , $AB = 6$ , $BC = 7$ , and $CA = 8$ . Point $D$ lies on $\overline{BC}$ , and $\overline{AD}$ bisects $\angle BAC$ . Point $E$ lies on $\overline{AC}$ , and $\overline{BE}$ bisects $\angle ABC$ . The bisectors intersect at $F$ . What is the ratio $AF$ : $FD$ ?

$美国数学竞赛$

$\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2$

Solution 1

Applying the angle bisector theorem to $\triangle ABC$ with $\angle CAB$ being bisected by $AD$ , we have

$\frac{CD}{AC}=\frac{BD}{AB}.$

Thus, we have

$amc美国数学竞赛真题$

and cross multiplying and dividing by $2$ gives us

$3\cdot CD=4\cdot BD.$

Since $CD+BD=BC=7$ , we can substitute $CD=7-BD$ into the former equation. Therefore, we get $3(7-BD)=4BD$ , so $BD=3$ .

Apply the angle bisector theorem again to $\triangle ABD$ with $\angle ABC$ being bisected. This gives us

$amc真题$

and since $AB=6$ and $BD=3$ , we have

$\frac{6}{AF}=\frac{3}{FD}.$

Cross multiplying and dividing by $3$ gives us

$AF=2\cdot FD,$

and dividing by $FD$ gives us

$amc12$

Therefore,

$美国数学竞赛真题$

Solution 2

By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$

$\frac{6}{AE} = \frac{7}{8 - AE}$ so $amc竞赛$

Similarly, $CD = 4$ .

Now, we use mass points. Assign point $C$ a mass of $1$ .

$mC \cdot CD = mB \cdot DB$ , so $mB = \frac{4}{3}$

Similarly, $A$ will have a mass of $美国数学竞赛$

$mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}$

So $amc数学竞赛$

Solution 3

Denote $[\triangle{ABC}]$ as the area of triangle ABC and let $r$ be the inradius. Also, as above, use the angle bisector theorem to find that $BD = 3$ . There are two ways to continue from here:

$1.$ Note that $F$ is the incenter. Then, $\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}$

$2.$ Apply the angle bisector theorem on $\triangle{ABD}$ to get $amc真题$

Solution 4

Draw the third angle bisector, and denote the point where this bisector intersects AB as P. Using angle bisector theorem, we see $AE=48/13 , EC=56/13, AP=16/5, PB=14/5$ . Applying Van Aubel's theorem, $AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1$ , and so the answer is $\boxed{\textbf{(C)}\; 2 : 1}$ .