2015 AIME II Problems/Problem 4

Problem

In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ .

Solution

Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$ , where $E$ is closer to $D$ .

Subtract the two bases and divide to find that $ED$ is $\log 8$ . The altitude can be expressed as $\frac{4}{3} \log 8$ . Therefore, the two legs are $\frac{5}{3} \log 8$ , or $\log 32$ .

The perimeter is thus $\log 32 + \log 32 + \log 192 + \log 3$ which is $\log 2^{16} 3^2$ . So $p + q = \boxed{018}$

Solution 2 (gratuitous wishful thinking)

Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\log 192 - \log 3 = \log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\log 64$ being 6. Then the 2 legs of the trapezoid is $\sqrt{3^2+4^2}=5=\log 32$ .

And we have the answer:

$\log 192 + \log 32 + \log 32 + \log 3 = \log(192 \cdot 32 \cdot 32 \cdot 3) = \log(2^6 \cdot 3 \cdot 2^5 \cdot 2^5 \cdot 3) = \log(2^{16} \cdot 3^2) \Rightarrow 16+2 = \boxed{18}$

-dragoon

Solution 3

Let $ABCD$ be the trapezoid, where $\overline{AB} || \overline{CD}$ and $AB = \log 3$ and $CD = \log 192$ . Draw altitudes from $A$ and $B$ to $\overline{CD}$ with feet at $E$ and $F$ , respectively. $AB = \log 3$ , so $EF = \log 3$ . Now, we attempt to find $DE + FC$ , or what's left of $CD$ after we take out $EF$ . We make use of the two logarithmic rules:

$\log(xy) = \log x + \log y$

$\log(x^a) = a\log(x)$

$CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2$

Thus, since $CD = DE + EF + FC = \log 3 + 6\log 2$ , $CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC$ .

Now, why was finding $DE + FC$ important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles $\triangle DAE$ and $\triangle BFC$ together to get $\triangle XC'D'$ , where $X$ is the point where $A$ and $B$ became one. Note we can do this because $\triangle DAE$ and $\triangle BFC$ are both right triangles with a common leg length (the altitude of trapezoid $ABCD$ ).

Triangle $XC'D'$ has a base of $C'D'%, which is just equal to$ (Error compiling LaTeX. Unknown error_msg)DE + FC = 6\log 2 $. It is equal to$ DE + FC $because when we brought triangles$ \triangle DAE $and$ \triangle BFC $together, the length of$ CD $was not changed except for taking out$ EF $.$ XC' = XD' $since$ AD = BC $because the problem tells us we have an isosceles trapezoid. Drop and altitude from$ X $to$ C'D' $The altitude has length$ \log 16 = 4\log 2 $. The altitude also bisects$ C'D' $since$ \triangle XC'D' $is isosceles. Let the foot of the altitude be$ M $. Then$ MD' = 3\log 2 $(Remember that C'D' was$ 6\log 2 $, and then it got bisected by the altitude). Thus, the hypotenuse,$ XD' $must be$ 5\log 2 $from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of$ \log 2 $. Since$ XD' = XC' = BC = AD $,$ BC = AD = 5\log 2 = \log 2^5$.

Now, we have$ (Error compiling LaTeX. Unknown error_msg)CD = \log (3 \cdot 2^6) $,$ AB = \log 3 $, and$ BC = AD = \log 2^5$. Thus, their sum is

Thus,$ (Error compiling LaTeX. Unknown error_msg)p + q = 16 + 2 = \boxed{18}$. ~Extremelysupercooldude

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=226s

~MathProblemSolvingSkills.com

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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