GPrep pb

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GPrep pb

by XIII » Mon Mar 31, 2008 7:14 pm
Having some trouble figuring out why P's coordinates don't mirror Q's. Thanks
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by tmmyc » Mon Mar 31, 2008 8:55 pm
This is what I posted in the Manhattan GMAT forums.
tmmyc wrote:First, see that after dropping perpendicular lines, we have two right triangles.

Image

Detailed Explanation:
Let's begin with the triangle on the left.

We know the sides are 1 and (sqrt 3) from point P.
If you know your special right triangles, you will quickly see that this is a 30-60-90 right triangle.

The angle opposite '1' is 30 degrees.



Let's move on to the triangle on the right.

We know that a straight line has 180 degrees.

Since we know the lower angle of the triangle on the left is 30 degrees, and we also know the angle between the two line segments is 90 degrees, the lower angle of the triangle on the right must be 60 degrees in order to sum to 180 degrees. (30 + 90 + x = 180; x = 60)

This means the triangle on the right is also a 30-60-90 triangle. The hypotenuse of this triangle is the same as the other triangle's (which is '2' by the Pythagorean Theorem), since both are radii of the same circle.

Using the same properties of a 30-60-90 triangle, you can find the side lengths and finally the point (s,t) which gives the value for s.

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by moneyman » Mon Mar 31, 2008 10:12 pm
This is the best explanation for this problem..There is so much logic into it!!
Maxx