## Monday, October 20, 2014

### 2014 AMC-8 prep : Test Date -- > Nov. 18th, 2014 -- Mathcounts minis that are harder/hardest AMC-8 level

Below are Mathcounts Minis for harder/the hardest AMC-8 problems.

I don't believe in cramming so no homework for this week.

However, if you can't have enough math and science, watch NOVA making more stuff series and start tinkering/making/ and working on some independent projects.

Mathcounts Minis :

New Approaches to Rate x Time = Distance

Algebra Rate

Finding Simpler Way to Approach a Complex Problem

More Constructive Counting

Areas of Irregular Convex Polygons

Seeing Patterns to Make Problems Easier Part I

Seeing Patterns to Make Problems Easier Part II

## Saturday, October 18, 2014

### 2015 Mathcounts State Prep: Mathcounts State Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.

Question: 2010 Mathcounts State Team Round #10: A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at (10, 0). The side of the square and the base of the triangle on the x-axis each equal 10 units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?

There are lots of similar triangles for this question, but I think this is the fastest way to find the area.
$$\Delta$$AGB is similar to $$\Delta$$DGC and their line ratio is 15 to 10 or 3 : 2.
$$\Delta$$CGF is similar to $$\Delta$$CBE.
$$\dfrac {CG} {CB}=\dfrac {GF} {BE}$$
$$\dfrac {2} {5}=\dfrac {GF} {10}$$$$\rightarrow GF=4$$ From there you get the area =
$$\dfrac{10\times 4} {2}=20$$

Question: 2010 Mathcounts State #30Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.
Since each side is the same for equilateral triangle ABC, once you use the 30-60-90 degree angle ratio and 45-45-90 degree angle ratio, you'll get the side.
Since area ratio stays constant, you can plug in any numbers and it's much easier to use integer first so I use 2 for $$\overline {CD}$$.
From there you get the side length for each side is $$\sqrt {3}+1$$.
$$\overline {AC}-C\overline {D}=\sqrt {3}+1-2$$ = $$\overline {AD} = \sqrt {3}-1$$
$$\Delta ABD$$ and $$\Delta CBD$$ share the same vertex, so their area ratio is just the side ratio, which is $$\dfrac {\sqrt {3}-1} {2}$$.