Difference between revisions of "User talk:Bobthesmartypants/Sandbox"
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</asy> | </asy> | ||
<cmath>\text{Find the probability that }b>a \text{.}</cmath> | <cmath>\text{Find the probability that }b>a \text{.}</cmath> | ||
+ | |||
+ | <asy>unitsize(2inch); | ||
+ | import olympiad; | ||
+ | path c2 = dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180); | ||
+ | path c1 = dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60); | ||
+ | path c3 = dir(120)..dir(150)..dir(180); | ||
+ | path c4 = dir(0)..dir(30)..dir(60); | ||
+ | |||
+ | draw(dir(0)..dir(30)..dir(60)..dir(90)..dir(120)..dir(150)..dir(180)--dir(0)); | ||
+ | draw(dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60)..dir(90)-dir(90)..dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180)--dir(90)-dir(0)); | ||
+ | draw((-1.2,0.8)--(1.2,0.8)); | ||
+ | label("$l_{-n}$", (1.2,0.8),dir(0)); | ||
+ | draw((-1.2,0.5)--(1.2,0.5)); | ||
+ | label("$l_0$", (1.2,0.5),dir(0)); | ||
+ | draw((-1.2,0.2)--(1.2,0.2)); | ||
+ | label("$l_n$", (1.2,0.2),dir(0)); | ||
+ | label("$\vdots$", (1.2, 0.4), dir(0)); | ||
+ | label("$\vdots$", (0, 0.4)); | ||
+ | label("$\vdots$", (1.2, 0.65), dir(0)); | ||
+ | label("$\vdots$", (0, 0.65)); | ||
+ | |||
+ | label("$A_{-n}$", intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); | ||
+ | label("$C_{-n}$", intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); | ||
+ | label("$D_{-n}$", intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); | ||
+ | label("$B_{-n}$", intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); | ||
+ | |||
+ | label("$X$", intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5)), dir(150)); | ||
+ | label("$Y$", intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5)), dir(30)); | ||
+ | |||
+ | label("$C_{n}$", intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); | ||
+ | label("$A_{n}$", intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); | ||
+ | label("$B_{n}$", intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); | ||
+ | label("$D_{n}$", intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); | ||
+ | |||
+ | dot(intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8))); | ||
+ | dot(intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8))); | ||
+ | dot(intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8))); | ||
+ | dot(intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8))); | ||
+ | |||
+ | dot(intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5))); | ||
+ | dot(intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5))); | ||
+ | dot(intersectionpoint(c3, (-1.2, 0.5)--(1.2, 0.5))); | ||
+ | dot(intersectionpoint(c4, (-1.2, 0.5)--(1.2, 0.5))); | ||
+ | |||
+ | dot(intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2))); | ||
+ | dot(intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2))); | ||
+ | dot(intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2))); | ||
+ | dot(intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2))); | ||
+ | </asy> | ||
+ | |||
+ | Two half-circles are drawn as shown above, with a line <math>l_0</math> throught the two intersections points, <math>X,Y</math> of the half-circles. Lines <math>l_k</math> for <math>k=-n\to n</math> parallel to the bases of the half-circles are drawn such that the distances between <math>l_k</math> and <math>l_0</math> and <math>l_{-k}</math> and <math>l_0</math> are always the same for all <math>k=1\to n</math>. | ||
+ | |||
+ | The intersection points of <math>l_k</math> with one of the half-circles are labeled <math>A_k, B_k</math>, and with the other half-circle at <math>C_k,D_k</math>, as shown in the diagram. | ||
+ | |||
+ | Prove that <cmath>\prod_{k=-n}^n |A_kB_k|+|C_kD_k| \ge \prod_{k=-n}^n |A_kD_k|+|B_kC_k|</cmath> | ||
+ | |||
==Picture 2== | ==Picture 2== | ||
<asy> | <asy> |
Latest revision as of 20:25, 6 November 2014
Contents
- 1 Bobthesmartypants's Sandbox
- 2 Solution 1
- 3 Solution 2
- 4 Picture 1
- 5 Picture 2
- 6 physics problem
- 7 Solution
- 8 inscribed triangle
- 9 moar images
- 10 yay
- 11 solution reflection
- 12 origami
- 13 combos
- 14 circles
- 15 more circles
- 16 checkerboasrd
- 17 Fermat point
- 18 cenn driagrma
- 19 cyclic square
- 20 diagram
- 21 Cyclic squares DOTS DTOS TDORS
Bobthesmartypants's Sandbox
Solution 1
First, continue to hit at . Also continue to hit at .
We have that . Because , we have .
Similarly, because , we have .
Therefore, .
We also have that because is a parallelogram, and .
Therefore, . This means that , so .
Therefore, .
Solution 2
Note that is rational and is not divisible by nor because .
This means the decimal representation of is a repeating decimal.
Let us set as the block that repeats in the repeating decimal: .
( written without the overline used to signify one number so won't confuse with notation for repeating decimal)
The fractional representation of this repeating decimal would be .
Taking the reciprocal of both sides you get .
Multiplying both sides by gives .
Since we divide on both sides of the equation to get .
Because is not divisible by (therefore ) since and is prime, it follows that .
Picture 1
Two half-circles are drawn as shown above, with a line throught the two intersections points, of the half-circles. Lines for parallel to the bases of the half-circles are drawn such that the distances between and and and are always the same for all .
The intersection points of with one of the half-circles are labeled , and with the other half-circle at , as shown in the diagram.
Prove that
Picture 2
physics problem
Solution
inscribed triangle
moar images
yay
solution reflection
origami
combos
circles
more circles
checkerboasrd
Fermat point
cenn driagrma
cyclic square
diagram
Cyclic squares DOTS DTOS TDORS