Difference between revisions of "G285 2021 Fall Problem Set"
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==Problem 2== | ==Problem 2== | ||
− | Let <math>\triangle ABC</math> be a right triangle with right angle at <math>B</math>, and <math>AC=12</math>. Let <math>D</math> denote the intersection of the cevian dropped from <math>B</math> onto <math>AC</math> such that <math>DA=DC</math>. If the reflection of point <math>B</math> across <math>D</math> lies on the circumcircle of <math>\triangle ABC</math> as <math>E</math>, | + | Let <math>\triangle ABC</math> be a right triangle with right angle at <math>B</math>, and <math>AC=12</math>. Let <math>D</math> denote the intersection of the cevian dropped from <math>B</math> onto <math>AC</math> such that <math>DA=DC</math>. If the reflection of point <math>B</math> across <math>D</math> lies on the circumcircle of <math>\triangle ABC</math> as <math>E</math>, <math>\sin(BAC)<\frac{5}{8}</math>, and the circumradius of <math>\triangle ABC</math> is an integer, find the smallest possible value of <math>AB^2+AE^2</math>. |
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+ | ==Problem 8== | ||
+ | Find <cmath>\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \sum_{d=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c+d)}}</cmath> | ||
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+ | <cmath>\textbf{(A)}\ \frac{16}{27} \qquad \textbf{(B)}\ \frac{32}{27} \qquad \textbf{(C)}\ \frac{64}{27} \qquad \textbf{(D)}\ \frac{128}{27} \qquad \textbf{(E)}\ \frac{256}{27}</cmath> | ||
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+ | [[G285 2021 Fall Problem Set Problem 8|Solution]] |
Latest revision as of 20:43, 18 July 2021
Welcome to the Fall Problem Set! There are problems, multiple-choice, and free-response.
Problem 1
Larry is playing a logic game. In this game, Larry counts , and removes the number for every th move, skipping for , and then increments by one. If starts at , what is when Larry counts his th integer? Assume
Problem 2
Let be a right triangle with right angle at , and . Let denote the intersection of the cevian dropped from onto such that . If the reflection of point across lies on the circumcircle of as , , and the circumradius of is an integer, find the smallest possible value of .
Problem 8
Find