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>. 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> | + | However, because of SSS congruency, <math>SHX'</math> must be congruent to <math>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>. | 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. | However, since this does not meet the requirements of the point being on land, it does not work. |
Revision as of 18:04, 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.