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

Revision as of 23:08, 21 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

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

Picture 2

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

sndbozx

$[asy] draw((9,0)--(21,0)--(16,12)--(0,12)--cycle); draw(Circle((12,6),6)); draw((12,0)--(12,12),linetype("8 8")+red); draw((12,6)--(228/13,108/13),linetype("8 8")+red); draw((12,6)--(9-27/(sqrt(313)),36/(sqrt(313))),linetype("8 8")+red); label("$A$",(0,12),N); label("$B$",(16,12),N); label("$C$",(21,0),S); label("$D$",(9,0),S); label("$E$",(12,12),N,red); label("$F$",(228/13,108/13),NE,red); label("$G$",(12,0),S,red); label("$H$",(9-27/(sqrt(313)),36/(sqrt(313))),SW,red); label("$O$",(12,6),NW); draw((16,12)--(16,0),linetype("8 8")+green); label("$P$",(16,0),S,green); [/asy]$ We are given that $AB=16$ , $CD=12$ , and $\overline{AB}\parallel\overline{CD}$ .

We can forget the restriction $BC<AD$ because if $BC>AD$ , we can just switch the labeling around so that $BC<AD$ .

Label the center of the inscribed circle $O$ ; Draw lines $\overline{OE}$ , $\overline{OF}$ , $\overline{OG}$ , and $\overline{OH}$

where $E$ , $F$ , $G$ , and $H$ are the tangent points of the circle and $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ and $\overline{DA}$ respectively.

Note that $\angle EBF+\angle GCF=180^{\circ}$ because the angles of quadrilateral $EGCB$ add up to $360^{\circ}$ , and $\angle OEB=\angle OGC=90^{\circ}$ .

Also, $\angle EBF+\angle EOF=180^{\circ}$ because the angles of quadrilateral $EBFO$ add up to $360^{\circ}$ , and $\angle OEB=\angle OFB=90^{\circ}$ .

Therefore, $\angle GCF=\angle EOF$ . By a similar reasoning, $\angle EBF=\angle GOF$ . Therefore, $EBFO\sim GOFC$ .

By a similar reasoning as the right side, $AEOH\sim OGDH$ .

Let $BE=a$ , and $AE=b$ , $DG=c$ , and $CG=d$ . We also know that $EO=FO=GO=HO=6$ because the diameter of circle $O$ is $12$ .

Since $EBFO\sim GOFC$ , then $\dfrac{c}{GO}=\dfrac{EO}{b}\implies\dfrac{c}{6}=\dfrac{6}{b}\implies bc=36$ .

Similarly, since $AEOH\sim OGDH$ , then $\dfrac{d}{GO}=\dfrac{EO}{a}\implies\dfrac{d}{6}=\dfrac{6}{a}\implies ad=36$ .

Also, $a+b=AB=16$ , and $c+d=CD=12$ . Therefore $b=16-a$ and $d=12-c$ .

We now substitute $b$ and $d$ into our other two equations: $(16-a)c=36$ and $a(12-c)=36$ .

Expanding gives $16c-ac=36$ and $12a-ac=36$ . Subtracting these two equations gives $12a-16c=0\implies 12a=16c\implies a=\dfrac{4}{3}c$ .

Substituting $a$ back into $a(12-c)=36$ yields $\dfrac{4}{3}c(12-c)=36\implies 16c-\dfrac{4}{3}c^2=36\implies \dfrac{4}{3}c^2-16c+36=0$

Solving this quadratic gives $c=3, 9$ . Based the the picture, $c$ is obviously not $9$ since $c<d$ , so $c=3$ . Therefore, $d=12-c=9$ .

(Note: if I had said $c=9$ , there wouldn't be any contradiction, apart from the fact that the picture would be drawn "flipped", i.e not to scale.)

Also, $a=\dfrac{4}{3}(3)=4$ and so $b=16-a=12$ .

All we need to find now is the length of $\overline{BC}$ . Draw the height $\overline{BP}$ with base $\overline{CD}$ .

Since $\angle BPC=90^{\circ}$ , we can use Pythagorean Theorem: $PC=d-a=5$ , $BP=12$ , therefore $BC=\sqrt{5^2+12^2}=\boxed{13}$ . $\Box$

@@ Line 75: / Line 75: @@
 </asy>
 <cmath>\text{Prove the shaded areas are equal.}</cmath>
+==sndbozx==
+<asy>
+draw((9,0)--(21,0)--(16,12)--(0,12)--cycle);
+draw(Circle((12,6),6));
+draw((12,0)--(12,12),linetype("8 8")+red);
+draw((12,6)--(228/13,108/13),linetype("8 8")+red);
+draw((12,6)--(9-27/(sqrt(313)),36/(sqrt(313))),linetype("8 8")+red);
+label("$A$",(0,12),N);
+label("$B$",(16,12),N);
+label("$C$",(21,0),S);
+label("$D$",(9,0),S);
+label("$E$",(12,12),N,red);
+label("$F$",(228/13,108/13),NE,red);
+label("$G$",(12,0),S,red);
+label("$H$",(9-27/(sqrt(313)),36/(sqrt(313))),SW,red);
+label("$O$",(12,6),NW);
+draw((16,12)--(16,0),linetype("8 8")+green);
+label("$P$",(16,0),S,green);
+</asy>
+We are given that <math>AB=16</math>, <math>CD=12</math>, and <math>\overline{AB}\parallel\overline{CD}</math>.
+We can forget the restriction <math>BC<AD</math> because if <math>BC>AD</math>, we can just switch the labeling around so that <math>BC<AD</math>.
+Label the center of the inscribed circle <math>O</math>; Draw lines <math>\overline{OE}</math>, <math>\overline{OF}</math>, <math>\overline{OG}</math>, and <math>\overline{OH}</math>
+where <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> are the tangent points of the circle and <math>\overline{AB}</math>, <math>\overline{BC}</math>, <math>\overline{CD}</math> and <math>\overline{DA}</math> respectively.
+Note that <math>\angle EBF+\angle GCF=180^{\circ}</math> because the angles of quadrilateral <math>EGCB</math> add up to <math>360^{\circ}</math>, and <math>\angle OEB=\angle OGC=90^{\circ}</math>.
+Also, <math>\angle EBF+\angle EOF=180^{\circ}</math> because the angles of quadrilateral <math>EBFO</math> add up to <math>360^{\circ}</math>, and <math>\angle OEB=\angle OFB=90^{\circ}</math>.
+Therefore, <math>\angle GCF=\angle EOF</math>. By a similar reasoning, <math>\angle EBF=\angle GOF</math>. Therefore, <math>EBFO\sim GOFC</math>.
+By a similar reasoning as the right side, <math>AEOH\sim OGDH</math>.
+Let <math>BE=a</math>, and <math>AE=b</math>, <math>DG=c</math>, and <math>CG=d</math>. We also know that <math>EO=FO=GO=HO=6</math> because the diameter of circle <math>O</math> is <math>12</math>.
+Since <math>EBFO\sim GOFC</math>, then <math>\dfrac{c}{GO}=\dfrac{EO}{b}\implies\dfrac{c}{6}=\dfrac{6}{b}\implies bc=36</math>.
+Similarly, since <math>AEOH\sim OGDH</math>, then <math>\dfrac{d}{GO}=\dfrac{EO}{a}\implies\dfrac{d}{6}=\dfrac{6}{a}\implies ad=36</math>.
+Also, <math>a+b=AB=16</math>, and <math>c+d=CD=12</math>. Therefore <math>b=16-a</math> and <math>d=12-c</math>.
+We now substitute <math>b</math> and <math>d</math> into our other two equations: <math>(16-a)c=36</math> and <math>a(12-c)=36</math>.
+Expanding gives <math>16c-ac=36</math> and <math>12a-ac=36</math>. Subtracting these two equations gives <math>12a-16c=0\implies 12a=16c\implies a=\dfrac{4}{3}c</math>.
+Substituting <math>a</math> back into <math>a(12-c)=36</math> yields <math>\dfrac{4}{3}c(12-c)=36\implies 16c-\dfrac{4}{3}c^2=36\implies \dfrac{4}{3}c^2-16c+36=0</math>
+Solving this quadratic gives <math>c=3, 9</math>. Based the the picture, <math>c</math> is obviously not <math>9</math> since <math>c<d</math>, so <math>c=3</math>. Therefore, <math>d=12-c=9</math>.
+(Note: if I had said <math>c=9</math>, there wouldn't be any contradiction, apart from the fact that the picture would be drawn "flipped", i.e not to scale.)
+Also, <math>a=\dfrac{4}{3}(3)=4</math> and so <math>b=16-a=12</math>.
+All we need to find now is the length of <math>\overline{BC}</math>. Draw the height <math>\overline{BP}</math> with base <math>\overline{CD}</math>.
+Since <math>\angle BPC=90^{\circ}</math>, we can use Pythagorean Theorem: <math>PC=d-a=5</math>, <math>BP=12</math>, therefore <math>BC=\sqrt{5^2+12^2}=\boxed{13}</math>. <math>\Box</math>

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