Difference between revisions of "User talk:Bobthesmartypants/Sandbox"
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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>. | 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>SHX'</math> must be congruent to <math>SHX</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> | 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> |
Revision as of 17:13, 8 December 2013
Bobthesmartypants's Sandbox
Solution 1
First, continue to hit at . Also continue to hit at .
We have that . Because , we have .
Similarly, because , we have .
Therefore, .
We also have that because is a parallelogram, and .
Therefore, . This means that , so .
Therefore, .
Solution 2
Note that is rational and is not divisible by nor because .
This means the decimal representation of is a repeating decimal.
Let us set as the block that repeats in the repeating decimal: .
( 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 .
Taking the reciprocal of both sides you get .
Multiplying both sides by gives .
Since we divide on both sides of the equation to get .
Because is not divisible by (therefore ) since and is prime, it follows that .
Picture 1
Picture 2
sandbox
PROVING THE EXISTENCE OF SUCH A POINT
We first want to prove that a point exists such that and .
First we draw a circle with center and radius . This denotes the locus of all points such that .
Now we draw a circle with center and radius . This denotes the locus of all points such that .
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: and .
We get rid of the extraneous solution, , because it does not satisfy the need that the treasure is on land.
Therefore the point that we seek is , and we have proved its existence.
Note that we are assuming that . This is true because the diagram given is to scale.
PROVING THAT THE POINT IS UNIQUE
Note that if there is another point , then it must satisfy that and .
We can let . The triangle therefore has sides of length , , and .
However, because of SSS congruency, must be congruent to .
Since , then and are the same point, and therefore is unique.
Note that there is an extraneous solution for that is to the left of the line .
However, since this does not meet the requirements of the point being on land, it does not work.