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Difference between revisions of "User talk:Bobthesmartypants/Sandbox"

(→‎Solution 2)
(→‎Solution 2)
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Because <math>n</math> is not divisible by <math>3</math> (therefore <math>9</math>) since <math>n>11</math> and <math>n</math> is prime, it follows that <math>n|\underbrace{111\cdots 111}_{x\text{ times}}</math>. <math>\Box</math>
 
Because <math>n</math> is not divisible by <math>3</math> (therefore <math>9</math>) since <math>n>11</math> and <math>n</math> is prime, it follows that <math>n|\underbrace{111\cdots 111}_{x\text{ times}}</math>. <math>\Box</math>
 +
==Picture 1==
 +
<asy>draw(Circle((1,1),2));
 +
draw(Circle((sqrt(2),sqrt(3)/2),1));
 +
dot((8/5,2/5));
 +
dot((1,1));
 +
draw((1,1)--(8/5,2/5),linetype("8 8"));
 +
label("a",(6/5,7/10),SSW);
 +
draw((8/5,2/5)--(12/5,-2/5),linetype("8 8"));
 +
label("b",(2,0),SSW);
 +
</asy>
 +
<cmath>\text{Find the probability that }b>a \text{.}</cmath>

Revision as of 20:10, 19 October 2013

Bobthesmartypants's Sandbox

Solution 1

[asy] path Q; Q=(0,0)--(1,2)--(5,2)--(4,0)--cycle; draw(Q); draw((0,0)--(1.5,1)); label("D",(0,0),S); draw((1,2)--(1.5,1)); label("A",(1,2),N); draw((5,2)--(1.5,1)); label("B",(5,2),N); draw((4,0)--(1.5,1)); label("C",(4,0),S); draw((2,0)--(1.5,1),linetype("8 8")); label("E",(2,0),S); draw((2/3,4/3)--(1.5,1),linetype("8 8")); label("F",(2/3,4/3),W); label("P",(1.5,1),NNE); [/asy]

First, continue $\overline{AP}$ to hit $\overline{CD}$ at $E$. Also continue $\overline{CP}$ to hit $\overline{AD}$ at $F$.

We have that $\angle PAB=\angle PCB$. Because $\overline{AB}\parallel\overline{CD}$, we have $\angle PAB=\angle PED$.

Similarly, because $\overline{AD}\parallel\overline{BC}$, we have $\angle PCB=\angle PFD$.

Therefore, $\angle PAB=\angle PED=\angle PCB=\angle PFD$.

We also have that $\angle ADC=\angle ABC$ because $ABCD$ is a parallelogram, and $\angle APC=\angle FPE$.

Therefore, $ABCP\sim FDEP$. This means that $\dfrac{FD}{AB}=\dfrac{FP}{AP}=\dfrac{DP}{BP}$, so $\Delta ABP\sim\Delta FDP$.

Therefore, $\angle PBA=\angle PDA$. $\Box$


Solution 2

Note that $\dfrac{1}{n}$ is rational and $n$ is not divisible by $2$ nor $5$ because $n>11$.

This means the decimal representation of $\dfrac{1}{n}$ is a repeating decimal.

Let us set $a_1a_2\cdots a_x$ as the block that repeats in the repeating decimal: $\dfrac{1}{n}=0.\overline{a_1a_2\cdots a_x}$.

($a_1a_2\cdots a_x$ 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 $\dfrac{1}{n}=\dfrac{a_1a_2\cdots a_x}{10^x-1}$.

Taking the reciprocal of both sides you get $n=\dfrac{10^x-1}{a_1a_2\cdots a_x}$.

Multiplying both sides by $a_1a_2\cdots a_n$ gives $n(a_1a_2\cdots a_x)=10^x-1$.

Since $10^x-1=9\times \underbrace{111\cdots 111}_{x\text{ times}}$ we divide $9$ on both sides of the equation to get $\dfrac{n(a_1a_2\cdots a_x)}{9}=\underbrace{111\cdots 111}_{x\text{ times}}$.

Because $n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\underbrace{111\cdots 111}_{x\text{ times}}$. $\Box$

Picture 1

[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype("8 8")); label("a",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype("8 8")); label("b",(2,0),SSW); [/asy] \[\text{Find the probability that }b>a \text{.}\]

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