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Tricky Geometry 2

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by GMATGuruNY » Wed Feb 01, 2012 4:54 am
Image

Square ABCD:
Since A = s² = 9/2, s = 3/√2.

QC:
To find the distance between 2 points -- in this case, the length of QC -- DRAW A RIGHT TRIANGLE so that the distance becomes the hypotenuse.
Drawing QE forms right triangle QCE.

Always look for SPECIAL TRIANGLES (45-45-90, etc) and for SIMILAR TRIANGLES.

Triangles DQE and DBC:
Since both are 45-45-90 triangles, they are similar.
Thus, all corresponding lengths are in the same ratio.
Since DQ:QB = 1:2, DQ = (1/3)DB.
Thus:
QE = (1/3)BC = (1/3)(3/√2) = 1/√2.
Since DE = (1/3)DC, EC = (2/3)DC = (2/3)(3/√2) = 2/√2.

Triangle QCE:
Since QE² + EC² = QC²:
QC = √( (1/√2)² + (2/√2)² )
= √ (1/2 + 2)
= √ (5/2)
= √5 / √2
= (√5*√2)/(√2*√2)
= √10/2.

The correct answer is A.
Last edited by GMATGuruNY on Wed Feb 01, 2012 8:08 am, edited 1 time in total.
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by MBACenter » Wed Feb 01, 2012 5:04 am
If the area of the square is 4.5, the length of each side is root(4.5).

Since the diagonal of a square is the hypotenuse of a 45-45-90 triangle, the length of DB is root(4.5) * root(2) = root(9) = 3.

Since QB is twice as long as DQ, we can say that the length of DQ is 1 and the length of QB is 2 (so it adds up to 3).

If we draw an imaginary line from Q down to DC perpendicular to DC and call the point where it intersects X, we will have another 45-45-90 triangle with bases DX and XQ and hypotenuse DC. DX and XQ will each measure 1/[root (2)] or [root (2)]/2.

We will also have another right triangle with bases XC and XQ and hypotenuse QC.

The measure of side XC is equal to DC - DX, or root(4.5) - 1/[root(2)]:

root(4.5) - 1/[root(2)]
root(9/2) - 1/[root(2)]
[root(9)]/[root(2)] - 1/[root(2)]
3/[root(2)] - 1/[root(2)]
2/[root(2)]
2[root(2)]/2
root(2)

So, by the Pythagorean theorem, we can find QC as follows:

(QC)^2 = [[root(2)]^2] + [([root (2)]/2)^2]
(QC)^2 = 2 + 2/4
(QC)^2 = 2 + 1/2
(QC)^2 = 5/2
QC = root(5/2)
QC = [root(5)]/[root(2)]
QC = [root(5)][root(2)]/2
QC = [root(10)]/2
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by shankar.ashwin » Wed Feb 01, 2012 5:24 am
Difficult problem to solve under 2 mins..

Couple of ways to eliminate wrong choices here..

We can find Side = 3/√2 from the area and diagonal will be 3.

In triangle BCD,any line(QC) dropped to the hypotenuse(BD) from C will be less than 3/√2 (the side)

At this point we can eliminate C,D and E (All > 3/√2)

The minimum length of a line (say Q`C dropped would be the altitude when BD is divided in 1:1 ratio)

In that case Q`C would be, Area of BCD = 1/2 * QC * BD

9/4 (Half of square) = 1/2 *Q`C*3

Q`C = 1.5

Now point Q is closer to the altitude point Q` than point D (side of square)

So QC would be closer to 1.5 than 3/√2.

Between A and B, only A works

No means an ideal solution, but we can easily eliminate CDE here.
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by GMATGuruNY » Wed Feb 01, 2012 8:44 am
shankar.ashwin wrote:Difficult problem to solve under 2 mins..
Nice job using the answer choices, Shankar!

While this problem is tricky, I'm not sure that its solution is especially time-consuming -- as long as we remember to apply a few common strategies:

-- making an unknown distance the hypotenuse of a right triangle
-- looking for special triangles and for similar triangles
-- applying the Pythagorean theorem

To clarify how these strategies are helpful here, I've fleshed out the reasoning in my solution above.
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