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solid geometry - Length, Pythagoras theorem

For COMPETITION
Number of Total Problems: 21.
FOR PRINT ::: (Book)

Problem Num : 21

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

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Point B is due east of point A. Point C is due north of point B. The distance between points A and C is $10sqrt 2$ , and $angle BAC= 45^circ$ . Point D is 20 meters due north of point C. The distance AD is between which two integers?

$extbf{(A)} 30 ext{and} 31 qquad extbf{(B)} 31 ext{and} 32 qquad extbf{(C)} 32 ext{and} 33 qquad extbf{...$
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Category Length, Pythagoras theorem

Analysis

Solution/Answer
By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.

Hence $AD=AF$ and $BD=BE$ and $CE=CF$ . Let $AD = x, BD = y$ and $CE = z$ gives three equations:
$x+y = a-1$
$x+z = a$
$y+z = a+1$
(where $a = 2012$ for the first triangle.)
Solving gives:
$x= frac{a}{2} - 1$
$y = frac{a}{2}$
$z = frac{a}{2}+1$
Subbing in gives that $T_2$ has sides of $1005, 1006, 1007$ .
$T_3$ can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with $a=1006$ ). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinately until the side lengths no longer form a triangle.
Subbing in gives $T_3$ with sides $502, 503, 504$ .
$T_4$ has sides $frac{501}{2}, frac{503}{2}, frac{505}{2}$ .
$T_5$ has sides $frac{499}{4}, frac{503}{4}, frac{507}{4}$ .
$T_6$ has sides $frac{495}{8}, frac{503}{8}, frac{511}{8}$ .
$T_7$ has sides $frac{487}{16}, frac{503}{16}, frac{519}{16}$ .
$T_8$ has sides $frac{471}{32}, frac{503}{32}, frac{535}{32}$ .
$T_9$ has sides $frac{439}{64}, frac{503}{64}, frac{567}{64}$ .
$T_{10}$ has sides $frac{375}{128}, frac{503}{128}, frac{631}{128}$ .
$T_{11}$ would have sides $frac{247}{256}, frac{503}{256}, frac{759}{256}$ but these length do not make a triangle as $frac{247}{256} + frac{503}{256} < frac{759}{256}$ .
Hence the perimeter is $frac{375}{128} + frac{503}{128} + frac{631}{128} = oxed{ extbf{(D)} frac{1509}{128}}$ $lacksquare$

Answer:

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