# 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 $\sqrt{3^2+4^2}=5=\log 32$. $\log 192 + \log 32 + \log 32 + \log 3 = \log(192 \cdot 32 \cdot 32 \cdot 3) = \log(2^6 \cdot 3 \cdot 2^5 \cdot 2^5 \cdot 3) = \log(2^{16} \cdot 3^2) \Rightarrow 16+2 = \boxed{18}$

-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 $DE + FC = 6\log 2$. 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 $CD = \log (3 \cdot 2^6)$, $AB = \log 3$, and $BC = AD = \log 2^5$. Thus, their sum is $$\log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)$$

Thus, $p + q = 16 + 2 = \boxed{18}$. ~Extremelysupercooldude

## Video Solution

~MathProblemSolvingSkills.com

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 