2014 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of an equilateral triangle

#1: P is the interior of equilateral triangle ABC, such that perpendicular segments from P to each of the sides of triangle ABC measure 2 inches, 9 inches and 13 inches. Find the number of square inches in the area of triangle ABC, and express your answer in simplest radical form.

#2: AMC 2007-B: Point P is inside equilateral  ABC. Points Q, R, and S are the feet of the
perpendiculars from P to AB, BC, and CA, respectively. Given that PQ = 1,PR = 2, and PS = 3, what is AB ?
#3:

This is an equilateral triangle. If the side is "S", the length of the in-radius would be\(\dfrac {\sqrt {3}} {6}\) of  S (or \(\dfrac {1} {3}\)of the height) and the length of the circum-radius would be\(\dfrac {\sqrt {3}} {3}\) of  S (or \(\dfrac {2} {3}\)of the height).

You can use 30-60-90 degree special right triangle angle ratio to get the length of each side as well as the height.

Solution I: Let the side be "s" and break the triangle into three smaller triangles.

\(\dfrac {9+13+2} {2}\)= 12s (base times height divided by 2)= \(\dfrac {\sqrt {3}} {4}\times s^{2}\)

s = 16√ 3 

Area of the triangle = 192√ 3



                                           
 

Solution II: Let the side be "s" and the height of the equilateral triangle be "h"

24* s (by adding 9, 2 and 13 since they are the height of each smaller triangle)  = s*h 
(Omit the divided by 2 part on either side since it cancels each other out.)

h = 24
Using 30-60-90 degree angle ratio, you get \(\dfrac {1} {2}\) s = 8√ 3  so  s = 16√ 3 

Area of the equilateral triangle = \(\dfrac {24\times 16\sqrt {3}} {2}\) = \(192\sqrt {3}\) 

#2: This one is similar to #1, the answer is 4√ 3 

Three Pole Problems : Similar triangles

Question: If you know the length of x and y, and the whole length of \(\overline {AB}\),

A: what is the ratio of a to b and 

B: what is the length of z.






 Solution for question A:
\(\Delta\)ABC and \(\Delta\)AFE are similar so \(\dfrac {z} {x}=\dfrac {b} {a+b}\). -- equation 1
Cross multiply and you have z ( a + b ) = bx

\(\Delta\)BAD and \(\Delta\)BFE are similar so \(\dfrac {z} {y}=\dfrac {a} {a+b}\). -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so \(\dfrac {x} {y}=\dfrac {a} {b}\)  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
\(\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}\)
\(\dfrac {zy+zx } {xy}=1\) \(\rightarrow\) z = \(\dfrac {xy} {x+y}\)

Applicable question: 

\(\overline {CD}=15\) and you know \(\overline {DB}:\overline {BC}=20:30=2:3\) 
 so  \(\overline {DB}=6\) and \(\overline {BC}=9\) 

\(\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }\) = 12

2014 Mathcounts State/National/AMC Prep : More Counting and Probability Problems from Mathcounts Mini / AoPS

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state prep

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

Probability with Geometry Representations  : Oh dear, the second half part is hilarious.

Probability with Geometry Representations : solution to the second half problem from previous video

Try this one from 1998 AIME #9. It's not too bad.

2014 Mathcounts State Prep : Folding Paper Questions

Folding paper questions are not too bad so here are two examples:
 

Question #1:  The two side lengths of rectangle ABCD is "a" and "b". If you fold along EF and the point B now converges on point D, what is the length of \(\overline {EF}\)?

Solution I :
Let the length of \(\overline {FC}\)  be x and the length of \(\overline {FD}\)  is thus a - x.

Using Pythagorean theorem you'll easily get x (the \( x^{2}\) part cancel each other out) and from there get the length of a - x.

\(\overline {HD}\) = \(\dfrac {1} {2}\) of the hypotenuse. (Use Pythagorean theorem or Pythagorean triples to get that length,)

Again, using Pythagorean theorem you'll get the length of \(\overline {HF}\). Times 2 to get \(\overline {EF}\).

Solution II : 
After you find the length of x, use \(\overline {EG}\), which is a - 2x and b as two legs of the right triangle EGF, you can easily get \(\overline {EF}\). (Pythagorean theorem)

Question #2: 

What about this time you fold B to touch the other side.
 What is the length of EF? 

This one is not too bad.

Do you see there are two similar triangles?

Just make sure you use the same corresponding sides to get the desired
length.

Answer to one mathleague quesiton from AoPS

Question is here.

You have two congruent triangles. 17-same angle and- x (SAS)
Using distance formula, the two green lines are of the same length.
\(\left( a-25\right) ^{2}+\left( 20-15\right) ^{2}=a^{2}+20^{2}\)
a = 5

Use another distance formula to get x -- the blue line.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)