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Difference between revisions of "2015 AIME II Problems/Problem 4"

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==Solution==
 
==Solution==
  
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Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. The point where an altitude intersects the larger base be <math>E</math> where <math>E</math> is closer to <math>D</math>.  
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Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. Let the point where an altitude intersects the larger base be <math>E</math>, where <math>E</math> is closer to <math>D</math>.  
  
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Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>.
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Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} \log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>.
  
 
The perimeter is thus <math>\log 32 + \log 32 + \log 192 + \log 3</math> which is <math>\log 2^{16} 3^2</math>. So <math>p + q = \boxed{018}</math>
 
The perimeter is thus <math>\log 32 + \log 32 + \log 192 + \log 3</math> which is <math>\log 2^{16} 3^2</math>. So <math>p + q = \boxed{018}</math>
  
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==Solution 2 (gratuitous wishful thinking)==
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Set the base of the log as 2. Then call the trapezoid <math>ABCD</math> with <math>CD</math> as the longer base. Then have the two feet of the altitudes be <math>E</math> and <math>F</math>, with <math>E</math> and <math>F</math> in position from left to right respectively. Then, <math>CF</math> and <math>ED</math> are <math>\log 192 - \log 3 = \log 64</math> (from the log subtraction identity. Then <math>CF=EF=3</math> (isosceles trapezoid and <math>\log 64</math> being 6. Then the 2 legs of the trapezoid is <math>\sqrt{3^2+4^2}=5=\log 32</math>.
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And we have the answer:
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<math>\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}</math>
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-dragoon
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==Solution 3==
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Let <math>ABCD</math> be the trapezoid, where <math>\overline{AB} || \overline{CD}</math> and <math>AB = \log 3</math> and <math>CD = \log 192</math>. Draw altitudes from <math>A</math> and <math>B</math> to <math>\overline{CD}</math> with feet at <math>E</math> and <math>F</math>, respectively. <math>AB = \log 3</math>, so <math>EF = \log 3</math>. Now, we attempt to find <math>DE + FC</math>, or what's left of <math>CD</math> after we take out <math>EF</math>. We make use of the two logarithmic rules:
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<cmath>\log(xy) = \log x + \log y</cmath>
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<cmath>\log(x^a) = a\log(x)</cmath>
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<cmath>CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2</cmath>
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Thus, since <math>CD = DE + EF + FC = \log 3 + 6\log 2</math>, <math>CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC</math>.
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Now, why was finding <math>DE + FC</math> important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles <math>\triangle DAE</math> and <math>\triangle BFC</math> together to get <math>\triangle XC'D'</math>, where <math>X</math> is the point where <math>A</math> and <math>B</math> became one. Note we can do this because <math>\triangle DAE</math> and <math>\triangle BFC</math> are both right triangles with a common leg length (the altitude of trapezoid <math>ABCD</math>).
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Triangle <math>XC'D'</math> has a base of <math>C'D'</math>, which is just equal to <math>DE + FC = 6\log 2</math>. It is equal to <math>DE + FC</math> because when we brought triangles <math>\triangle DAE</math> and <math>\triangle BFC</math> together, the length of <math>CD</math> was not changed except for taking out <math>EF</math>.
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<math>XC' = XD'</math> since <math>AD = BC</math> because the problem tells us we have an isosceles trapezoid. Drop and altitude from <math>X</math> to <math>C'D'</math> The altitude has length <math>\log 16 = 4\log 2</math>. The altitude also bisects <math>C'D'</math> since <math>\triangle XC'D'</math> is isosceles. Let the foot of the altitude be <math>M</math>. Then <math>MD' = 3\log 2</math> (Remember that C'D' was <math>6\log 2</math>, and then it got bisected by the altitude). Thus, the hypotenuse, <math>XD'</math> must be <math>5\log 2</math> from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of <math>\log 2</math>. Since <math>XD' = XC' = BC = AD</math>, <math>BC = AD = 5\log 2 = \log 2^5</math>.
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Now, we have <math>CD = \log (3 \cdot 2^6)</math>, <math>AB = \log 3</math>, and <math>BC = AD = \log 2^5</math>. Thus, their sum is
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<cmath> \log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)</cmath>
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Thus, <math>p + q = 16 + 2 = \boxed{18}</math>. ~Extremelysupercooldude
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==Video Solution==
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https://www.youtube.com/watch?v=9re2qLzOKWk&t=226s
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~MathProblemSolvingSkills.com
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==See also==
 
{{AIME box|year=2015|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2015|n=II|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:17, 28 June 2023

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 $DE + FC = 6\log 2$. 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

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

~MathProblemSolvingSkills.com


See also

2015 AIME II (Problems • Answer Key • Resources)
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Problem 5
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