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

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

sandbox

$[asy] unitsize(0.2mm); pair H,S,X,Y,A,B; H = (25,0); S = (0,115); X = (122,26); Y = (-75,-17); A = ((H+X)/2); B = ((S+X)/2); draw(Circle(H,100)); draw(Circle(S,150)); draw(H--S--X--cycle); draw(H--Y--S,linetype("8 8")); label("100",A,dir(-90)); label("150",B,dir(-120)); label("H",H,dir(-90)); label("S",S,dir(90)); label("X",X,dir(0)); label("X'",Y,dir(-135)); [/asy]$

PROVING THE EXISTENCE OF SUCH A POINT

We first want to prove that a point $X$ exists such that $HX=100$ and $SX=150$ .

First we draw a circle with center $H$ and radius $100$ . This denotes the locus of all points $P$ such that $HP=100$ .

Now we draw a circle with center $S$ and radius $150$ . This denotes the locus of all points $P$ such that $SP=150$ .

Note that the intersection of these two locuses are the points which satisfy both conditions.

We see that there are two points which satisfy both locuses: $X$ and $X'$ .

We get rid of the extraneous solution, $X'$ , because it does not satisfy the need that the treasure is on land.

Therefore the point that we seek is $X$ , and we have proved its existence. $\Box$

Note that we are assuming that $SH\le SX+HX$ . This is true because the diagram given is to scale.

PROVING THAT THE POINT IS UNIQUE

$[asy] unitsize(0.2mm); pair H,S,X,Y,A,B,C; H = (25,0); S = (0,115); X = (122,26); Y = (120,50); A = ((H+X)/2); B = ((S+X)/2); C = ((H+S)/2); draw(H--S--X--cycle); draw(H--Y--S,linetype("8 8")); label("100",A,dir(-90)); label("150",B,dir(-120)); label("H",H,dir(-90)); label("S",S,dir(90)); label("X",X,dir(0)); label("X'",Y,dir(45)); label("n",C,dir(210)); [/asy]$

Note that if there is another point $X'$ , then it must satisfy that $HX'=100$ and $SX'=150$ .

We can let $SH=n$ . The triangle $SHX'$ therefore has sides of length $n$ , $100$ , and $150$ .

However, because of SSS congruency, $SHX'$ must be congruent to $SHX$ . Since $SX'=SX$ , then $X$ and $X'$ are the same point, and therefore $X$ is unique. $\Box$

Note that there is an extraneous solution for $X'$ that is to the left of the line $\overline{SH}$ .

However, since this does not meet the requirements of the point being on land, it does not work.

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