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Questions from the last mathelete of 2007-2008

Posted February 6th, 2008 by Mgccl
in

No more matheletes Crying. I don't wana... but we will have a pizza party so, it's still cool. Running Happily
Have to post them now or else I will forget.

A inscribed square in a circle have an area of 60, what is the area of the square circumscribe the circle?

The inscribed square with area of 60, each side of the square is x, then
x^2=60
and the diameter of the circle is
\sqrt{x^2+x^2} = \sqrt{120}
the area of the square circumscribe the circle is the diameter of the circle squared, answer is 120.

log_{x}\sqrt{x+12}>1, find all real x.

\begin{eqnarray} log_{x}\sqrt{x+12}&>&1\\ \sqrt{x+12}&>&x^1\\ x+12&>&x^2\\ 0&>&x^2-x-12\\ \end{eqnarray}
Find x = {4, -3}
but there is no log for negative base, so x<4
x>1 because log base one result one, and log base between 0 and one result negative number when the number getting logged is larger than the base.
1<x<4

find the last digit of 0!+1! + 2! + 3!....+20!

5! contains a 10, so x! end with 0 when x >= 5.
now, just sum up last digit of 0! + 1! + 2! + 3! + 4!, which is 4

A trapezoid have 2 bases with length 10 and 16, and 2 legs length 8 and 10, what is the area of this trapezoid.

separate this trapezoid into a rectangle and one triangle. 16-10 = 6, result a 6,8,10 triangle, it's easy to see it's a right angle triangle, and 8 is the height of the trapezoid. The answer is (16+10)*8/2 = 104

Square ABCD, E is the middle point of AB. O is the intersection of BD and EC. BO = 2. What's the area of this square.

Triangle EOB is similar to triangle COD because angle EOB congruent to angle DOC, angle EBO congruent to angle ODC. OD = 2 * BO = 4 since EB = 1/2DC. DB = 2+4 = 6. Just solve for x^2 in \sqrt{x^2+x^2} = 6, result is 18

\sqrt{5x} - \sqrt{2x} = 5 - 2, solve for x

I really didn't get why the 5-2 when they can use 3. I believe it's use to confuse people.
\begin{eqnarray} \sqrt{5x} - \sqrt{2x} &=& 5 - 2\\ (\sqrt{5}-\sqrt{2})(\sqrt{x}) &=& 3\\ \sqrt{x} &=& \frac{3}{\sqrt{5}-\sqrt{2}}\\ x &=& \frac{9}{7-2\sqrt{10}}\\ x &=& \frac{9(7+2\sqrt{10})}{(7-2\sqrt{10})(7+2\sqrt{10})}\\ x &=& 7+2\sqrt{10}\\ \end{eqnarray}

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