Difference between revisions of "2015 AIME II Problems/Problem 4"
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+ | ==Solution 3== | ||
+ | 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: | ||
+ | <cmath>\log(xy) = \log x + \log y</cmath> | ||
+ | |||
+ | <cmath>\log(x^a) = a\log(x)</cmath> | ||
+ | |||
+ | <cmath>CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>). | ||
+ | |||
+ | Triangle <math>XC'D'</math> has a base of <math>C'D'%, 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>. | ||
+ | |||
+ | </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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <cmath> \log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)</cmath> | ||
+ | |||
+ | Thus, </math>p + q = 16 + 2 = \boxed{18}$. ~Extremelysupercooldude | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:16, 28 June 2023
Contents
[hide]Problem
In an isosceles trapezoid, the parallel bases have lengths and , and the altitude to these bases has length . The perimeter of the trapezoid can be written in the form , where and are positive integers. Find .
Solution
Call the trapezoid with as the smaller base and as the longer. Let the point where an altitude intersects the larger base be , where is closer to .
Subtract the two bases and divide to find that is . The altitude can be expressed as . Therefore, the two legs are , or .
The perimeter is thus which is . So
Solution 2 (gratuitous wishful thinking)
Set the base of the log as 2. Then call the trapezoid with as the longer base. Then have the two feet of the altitudes be and , with and in position from left to right respectively. Then, and are (from the log subtraction identity. Then (isosceles trapezoid and being 6. Then the 2 legs of the trapezoid is .
And we have the answer:
-dragoon
Solution 3
Let be the trapezoid, where and and . Draw altitudes from and to with feet at and , respectively. , so . Now, we attempt to find , or what's left of after we take out . We make use of the two logarithmic rules:
Thus, since , .
Now, why was finding important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles and together to get , where is the point where and became one. Note we can do this because and are both right triangles with a common leg length (the altitude of trapezoid ).
Triangle has a base of $C'D'%, which is just equal to$ (Error compiling LaTeX. Unknown error_msg)DE + FC = 6\log 2DE + FC\triangle DAE\triangle BFCCDEFXC' = XD'AD = BCXC'D'\log 16 = 4\log 2C'D'\triangle XC'D'MMD' = 3\log 26\log 2XD'5\log 2\log 2XD' = XC' = BC = ADBC = AD = 5\log 2 = \log 2^5$.
Now, we have$ (Error compiling LaTeX. Unknown error_msg)CD = \log (3 \cdot 2^6)AB = \log 3BC = AD = \log 2^5$. Thus, their sum is
<cmath> \log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)</cmath>
Thus,$ (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
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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