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考题20-21 2015 AMC 10A

2018-08-07 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数学竞赛真题以及官方解答吧：

Problem 20

A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$ ?

$考题20-21 2015 AMC 10A$

Solution

Let the rectangle's length be $a$ and its width be $b$ . Its area is $ab$ and the perimeter is $2a+2b$ .

Then $A + P = ab + 2a + 2b$ . Factoring, we have $考题20-21 2015 AMC 10A$ .

The only one of the answer choices that cannot be expressed in this form is $102$ , as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$ , keeping $a$ and $b$ as positive integers.

Our answer is then $\boxed{B}$ .

Note: The original problem only stated that $A$ and $P$ were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.

Problem 21

Tetrahedron $ABCD$ has $AB=5$ , $AC=3$ , $BC=4$ , $考题20-21 2015 AMC 10A$ , $AD=3$ , and $CD=\tfrac{12}5\sqrt2$ . What is the volume of the tetrahedron?

$考题20-21 2015 AMC 10A$

Solution 1

Let the midpoint of $CD$ be $E$ . We have $CE = \dfrac{6}{5} \sqrt{2}$ , and so by the Pythagorean Theorem $AE = \dfrac{\sqrt{153}}{5}$ and $BE = \dfrac{\sqrt{328}}{5}$ . Because the altitude from $A$ of tetrahedron $ABCD$ passes touches plane $BCD$ on $BE$ , it is also an altitude of triangle $ABE$ . The area $A$ of triangle $ABE$ is, by Heron's Formula, given by

$考题20-21 2015 AMC 10A$ Substituting $a = AE, b = BE, c = 5$ and performing huge (but manageable) computations yield $A^2 = 18$ , so $A = 3\sqrt{2}$ . Thus, if $h$ is the length of the altitude from $A$ of the tetrahedron, $BE \cdot h = 2A = 6\sqrt{2}$ . Our answer is thus $V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},$ and so our answer is $\boxed{\textbf{(C) } \dfrac{24}{5}}$

Solution 2

Drop altitudes of triangle $ABC$ and triangle $ABD$ down from $C$ and $D$ , respectively. Both will hit the same point; let this point be $T$ . Because both triangle $ABC$ and triangle $ABD$ are 3-4-5 triangles, $CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}$ . Because $CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}$ , it follows that the $CTD$ is a right triangle, meaning that $\angle CTD = 90^\circ$ , and it follows that planes $ABC$ and $ABD$ are perpendicular to each other. Now, we can treat $ABC$ as the base of the tetrahedron and $TD$ as the height. Thus, the desired volume is $V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}$ which is answer $amc数学竞赛真题$