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Showing posts with label 2014 Mathcounts state prep. Show all posts
Showing posts with label 2014 Mathcounts state prep. Show all posts

Friday, January 20, 2023

2015 Mathcounts State Prep: Mathcounts State Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

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.
  ΔAGB is similar to ΔDGC and their line ratio is 15 to 10 or 3 : 2.
     ΔCGF is similar to ΔCBE. 
     CGCB=GFBE
     25=GF10GF=4 From there you get the area =
      10×42=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 CD.
From there you get the side length for each side is 3+1.
ACCD=3+12 = AD=31
ΔABD and ΔCBD share the same vertex, so their area ratio is just the side ratio, which is 312.

Wednesday, October 5, 2016

2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded

How to find volume of a tetrahedron (right pyramid) with side length one.

The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).


The side of the cube is S2 so the volume of the regular embedded tetrahedron is
13×(S2)3=13s322= 2S312.

You can also fine the height of the tetrahedron and then 13*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is 23 of the height of the equilateral triangle base, you'll get the space height.

Monday, February 17, 2014

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 be36 of  S (or 13of the height) and the length of the circum-radius would be33 of  S (or 23of 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.


9+13+22= 12s (base times height divided by 2)= 34×s2

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 12 s = 8 3  so  s = 16 3 
Area of the equilateral triangle = 24×1632 = 1923 

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


Thursday, January 23, 2014

Three Pole Problems : Similar triangles






Question: If you know the length of x and y, and the whole length of AB,

A: what is the ratio of a to b and 

B: what is the length of z.






 Solution for question A:
ΔABC and ΔAFE are similar so zx=ba+b. -- equation 1
Cross multiply and you have z ( a + b ) = bx

ΔBAD and ΔBFE are similar so zy=aa+b. -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so xy=ab  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
zx+zy=ba+b+aa+b
zy+zxxy=1 z = xyx+y




Applicable question: 

CD=15 and you know DB:BC=20:30=2:3 
 so  DB=6 and BC=9 

AB=20×30(20+30) = 12