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

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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

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

Picture 2

$[asy] for (int i=0;i<6;i=i+1){ draw(dir(60*i)--dir(60*i+60)); } draw(dir(120)--(dir(0)+dir(-60))/2); draw(dir(180)--(dir(60)+dir(0))/2); fill(dir(120)--dir(180)--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); fill((dir(0)+dir(-60))/2--dir(0)--(dir(60)+dir(0))/2--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); [/asy]$ $\text{Prove the shaded areas are equal.}$ $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,black); } } for(int i = 1; i < 7; ++i){ filldraw((i,7-i)--(i+1,7-i)--(i+1,8-i)--(i,8-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((i,4-i)--(i+1,4-i)--(i+1,5-i)--(i,5-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((8-i,4+i)--(7-i,4+i)--(7-i,3+i)--(8-i,3+i)--cycle,white); } filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle,white); filldraw((0,1)--(0,2)--(1,2)--(1,1)--cycle,white); filldraw((7,8)--(6,8)--(6,7)--(7,7)--cycle,white); filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white); [/asy]$

@@ Line 97: / Line 97: @@
 filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white);
 </asy>
+==physics problem==

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