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

Revision as of 18:30, 25 April 2014

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 75: / Line 75: @@
 </asy>
 <cmath>\text{Prove the shaded areas are equal.}</cmath>
-==sandbox==
 <asy>
-unitsize(0.2mm);
+for(int i = 0; i < 8; ++i){
-pair H,S,X,Y,A,B;
+  for(int j = 0; j < 8; ++j){
-H = (25,0);
+    filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,black);
-S = (0,115);
+  }
-X = (122,26);
+}
-Y = (-75,-17);
+for(int i = 1; i < 7; ++i){
-A = ((H+X)/2);
+  filldraw((i,7-i)--(i+1,7-i)--(i+1,8-i)--(i,8-i)--cycle,white);
-B = ((S+X)/2);
+}
-draw(Circle(H,100));
+for(int i = 0; i < 5; ++i){
-draw(Circle(S,150));
+  filldraw((i,4-i)--(i+1,4-i)--(i+1,5-i)--(i,5-i)--cycle,white);
-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 <math>X</math> exists such that <math>HX=100</math> and <math>SX=150</math>.
-First we draw a circle with center <math>H</math> and radius <math>100</math>. This denotes the locus of all points <math>P</math> such that <math>HP=100</math>.
-Now we draw a circle with center <math>S</math> and radius <math>150</math>. This denotes the locus of all points <math>P</math> such that <math>SP=150</math>.
-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: <math>X</math> and <math>X'</math>.
-We get rid of the extraneous solution, <math>X'</math>, because it does not satisfy the need that the treasure is on land.
-Therefore the point that we seek is <math>X</math>, and we have proved its existence. <math>\Box</math>
-Note that we are assuming that <math>SH\le SX+HX</math>. This is true because the diagram given is to scale.
-'''PROVING THAT THE POINT IS UNIQUE'''
+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);
+}
-<asy>
+filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle,white);
-unitsize(0.2mm);
+filldraw((0,1)--(0,2)--(1,2)--(1,1)--cycle,white);
-pair H,S,X,Y,A,B,C;
+filldraw((7,8)--(6,8)--(6,7)--(7,7)--cycle,white);
-H = (25,0);
+filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white);
-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 <math>X'</math>, then it must satisfy that <math>HX'=100</math> and <math>SX'=150</math>.
-We can let <math>SH=n</math>. The triangle <math>SHX'</math> therefore has sides of length <math>n</math>, <math>100</math>, and <math>150</math>.
-However, because of SSS congruency, <math>\Delta SHX'</math> must be congruent to <math>\Delta SHX</math>.
-Since <math>SX'=SX</math>, then <math>X</math> and <math>X'</math> are the same point, and therefore <math>X</math> is unique. <math>\Box</math>
-Note that there is an extraneous solution for <math>X'</math> that is to the left of the line <math>\overline{SH}</math>.
-However, since this does not meet the requirements of the point being on land, it does not work.

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