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考题20 2015 AMC真题 12A

2018-08-22 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 20

Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ , $5$ , and $8$ , while those of $T'$ have lengths $a$ , $a$ , and $b$ . Which of the following numbers is closest to $b$ ?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution

Solution 1

The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.

The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$ .

Thus $2a + b = 18$ , so $2a = 18 - b$ .

Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ , so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$ .

We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$ , so $b^3 - 9b^2 + 64 = 0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\textbf{(A)}$ .

Solution 1.1

The area is $12$ , the semiperimeter is $9$ , and $a = 9 - \frac12b$ . Using Heron's formula, $amc数学竞赛官网$ . Squaring both sides and simplifying, we have $-b^3+9b-64=0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\textbf{(A)}$ .

Solution 2

Triangle $T$ , being isosceles, has an area of $\frac{1}{2}(8)\sqrt{5^2-4^2}=12$ and a perimeter of $5+5+8=18$ . Triangle $T'$ similarly has an area of $\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ and $2a+b=18$ .

Now we apply our computational fortitude.

$\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ $AMC12数学竞赛试题及答案$ $(b)\sqrt{4a^2-b^2}=48$ $b^2(4a^2-b^2)=48^2$ $AMC12数学竞赛试题及答案$ Plug in $2a+b=18$ to obtain $18b^2(2a-b)=48^2$ $b^2(2a-b)=128$ Plug in $2a=18-b$ to obtain $b^2(18-2b)=128$ $2b^3-18b^2+128=0$ $b^3-9b^2+64=0$ We know that $b=8$ is a valid solution by $T$ . Factoring out $b-8$ , we obtain $(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0$ Utilizing the quadratic formula gives $b=\frac{1\pm\sqrt{33}}{2}$ We clearly must pick the positive solution. Note that $5<\sqrt{33}<6$ , and so ${3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}$ , which clearly gives an answer of $\fbox{A}$ , as desired.

Solution 3

Triangle T has perimeter $5 + 5 + 8 = 18$ so $18 = 2a + b$ .

Using Heron's, we get $\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})(\frac{b}{2})^2(\frac{2a-b}{2})}$ .

We know that $2a + b = 18$ from above so we plug that in, and we also know that then $2a - b = 18 - 2b$ .

$12 = 3\frac{b}{2}\sqrt{9-b}$

$64 = 9b^2 - b^3$

We plug in 3 for $b$ in the LHS, and we get 54 which is too low. We plug in 4 for $b$ in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4.

If $b \geq 3.5$ , then we would round up to 4, but if $b < 3.5$ , then we would round down to 3. So let us plug in 3.5 for b.

We get 67.375 which is too high, so we know that $b < 3.5$ .

The answer is $3$ . $\textbf{(A)}$

Operation Descartes

For this new triangle, say its legs have length $d$ and the base length $2c$ . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that $c+d=9$ and $c*\sqrt{d^2-c^2}=12$ . It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!

Now, modify the square-root equation with $d=9-c$ ; you get $c^2*(81-18c)=144$ , so $-18c^3+81c^2=144$ . Divide by $-9$ to get $2c^3-9c^2+16=0$ . Obviously, $c=4$ is a root as established by triangle $T$ ! So, use synthetic division to obtain $2c^2-c-4=0$ , upon which $c=\frac{1+\sqrt{33}}{4}$ , which is closest to $\frac{3}{2}$ (as opposed to $2$ ). That's enough to confirm that the answer has to be $\textbf{A}$ .