Harder Mathcounts State/AMC Questions: Intermediate level if you can solve in less than 2 mins.

2012 Mathcounts State Sprint #30: In rectangle ABCD, shown here, point M is the midpoint of side BC, and point N lies on CD such that DN:NC = 1:4. Segment BN intersects AM and AC at points R and S, respectively. If NS:SR:RB = x:y:z, where x, y and z are positive integers, what is the minimum possible value of x + y + z? 

Solution I :

\(\overline {AB}:\overline {NC}=5:4\) [given]

Triangle ASB is similar to triangle CSN (AAA)

\(\overline {NS}:\overline {SB}= 4 : 5\)

Let \(\overline {NS}= 4a,  \overline {SB}= 5a.\)

Draw a parallel line to \(\overline {NC}\) from M and mark the interception to \(\overline {BN}\)as T.

 \(\overline {MT}: \overline {NC}\) = 1 to 2. [\(\Delta BMT\) and \(\Delta BCN\) are similar triangles ]

\(\overline {NT} = \overline {TB}= \dfrac {4a+5a} {2}=4.5a\)

\(\overline {ST} = 0.5a\)

 \(\overline {MT} :  \overline {AB}\) = 2 to 5
[Previously we know  \(\overline {MT}: \overline {NC}\) = 1 to 2 or 2 to 4 and  \(\overline {NC}:\overline {AB}= 4 : 5\) so the ratio of the two lines  \(\overline {MT} :  \overline {AB}\) is 2 to 5.]


\(\overline {TB} = 4.5 a\)  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles \(\Delta ARB\) and \(\Delta MRT\) , you get \(\overline {BR} =\dfrac {5} {7}\times 4.5a =\dfrac {22.5a} {7}\) and \(\overline {RT} =\dfrac {2} {7}\times 4.5a =\dfrac {9a} {7}\)

Thus, x : y : z = 4a : \( \dfrac {1} {2}a + \dfrac {9a} {7}\) : \(\dfrac {22.5a} {7}\) = 56 : 25 : 45

x + y + z = 126

Solution II : 
From Mathcounts Mini: Similar Triangles and Proportional Reasoning

Solution III: 
Using similar triangles ARB and CRN , you have \(\dfrac {x} {y+z}=\dfrac {5} {9}\).
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have \( \dfrac {x+y} {z}=\dfrac {5} {4}\).
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = \(\dfrac {9} {5}y\)
Plug in x = \(\dfrac {9} {5}y\) to equation II and you have z  =  \( \dfrac {56} {25}y\)

x : y : z = \(\dfrac {9} {5}y\)  : y  :  \( \dfrac {56} {25}y\) =  45 y :  25y  :  56y

45 + 25 + 56 = 126

Solution IV : Yes, there is another way that I've found even faster, saved for my private students. :D 

Solution V : from Abhinav, one of my students solving another similar question : 

Two other similar questions from 2016 AMC A, B tests : 

2016 AMC 10 A, #19 : Solution from Abhinav 

2016 AMC 10 B #19 : Solution from Abhinav 

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.
  \(\Delta \)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 #30:  Point 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 \(\dfrac {\sqrt {3}-1} {2}\).

Mass Points Geometry

Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Basics 

2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40

(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Another useful notes 

Videos on Mass Point :

Mass Points Geometry Part I 

Mass Points Geometry : Split Masses Part II 

Mass Points Geometry : Part III 

other videos from Youtube on Mass Points

It's much more important to fully understand how it works, the easier questions the weights align
very nicely.

The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.

Let me know if you have questions. I love to help (:D) if you've tried.

Geometric mean of a right triangle

Geometric mean of a right triangle :

Video tutorial 

Notes

Practice I 

Practice II : focus on the height (or altitude)

Practice III : focus on the legs 

The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

Proof without words from Mr. Rusczyk 

Try using different types of triangles to experiment and see for yourself.
Paper folding is fun !!!!!
It's very cool :D

Original Problem from a Student Problem Solver on Similar Triangles

An original problem on similar triangles from Varun (from FL)

Square Based Pyramid of Equal Edges: Vomume and Surface Area

The height of a square based pyramid of equal edges is half of its base diagonal. 

You should be able to figure out the volume as well as surface area of a square based pyramid once 

you understand how to get the slant height and the height. 

2013 Mathcounts Nationals Sprint #29 and similar problem hints

Mathcounts problems : state/nationals level

2013 Mathcounts Natinals Sprint # 28

2013 Mathcounts National Sprint #28 : In right triangle ABC, shown here,  line AC = 5 units and line BC = 12 units. Points D and E lie on  line AB and line BC respectively, so that line CD is perpendicular to line AB and E is the midpoint of line BC. Segments AE and CD intercept at point F. What is the ratio of AF to FE ? Express your answer as a common fraction.

                                                     Solution I : Using similar triangles

                                                        Solution II : Use Mass Point Geometry

Geometry : Harder Chapter Level Quesitons

Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 √ 3 . What is the area of the a. smaller triangle   b. larger triangle?  

Solution: 
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be √ 4  to √ 9  
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 √ 3 so the smaller triangle has a perimeter of
2/5 *  30√ 3 or 12 √ 3. One side is 4√ 3 . Using the formula of finding the area of an equilateral triangle \(\dfrac{\sqrt3*s^2}{4}\) , you get the area to be 12√ 3.

Use the same method to get the area of the larger triangle as 27√ 3.
You can also use ratio relationship to get the area of the larger triangle by 
multiply 12√ 3 by 9/4.


2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 8√2, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]

Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
\(\dfrac{(16+x)* (16-x)}{4}\) = 48 ; 256 - x² = 192 ;  - x² = - 64;  x = 8 = YZ

Solution II: 

Make the y be the height of the trapezoid. YZ = 16 - 2y.  \(\dfrac{(16-2y + 16)}{2}\) * y = 48
\({(16 -y)* y = 48}\)\(\rightarrow\) \({16y -y^2 = 48}\) \(\rightarrow\) \({y^2 - 16y + 48 = 0}\) \(\rightarrow\) \({(y -12)(y -4)=0}\) \(\rightarrow\) \({y = 4}\) or \({y = 12}\)(doesn't work)
 YZ = 16 - 2y. Plug in y = 4 and you have  YZ = 8

Solution III: Let the height be y and you have \(\dfrac{(\overline{YZ}+ 16)* y}{2}"\) = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8

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 \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).

You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.

Mathcounts : Geometry -- Medians ; Squares in Isosceles Right Triangle, Similar Triangles, Triangles share the same vertex

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

This one appears at 91 Mathcounts National Target #4. It's interesting.

Q: triangle ABC is equilateral and G is the midpoint. BI is the same length as BC and the question asks about the area of quadrilateral polygon GDBC.
We'll also find the area of triangle ADG and triagle DBI here. 

Solution I: The side length is 2 and it's an equilateral triangle so the area of triangle ABC is 
\(\dfrac {\sqrt {3}} {4}\times 2^{2}=\sqrt {3}\).
Using similar triangles GCF and ACE, you get \(\overline {GF}\)= \(\dfrac {\sqrt {3}} {2}\)
From there, you get area of  \(\Delta\)GIC = \(4*\dfrac {\sqrt {3}} {2}*\dfrac {1} {2}=\sqrt {3}\)
 \(\Delta\)FCH is a 30-60-90 degree right triangle so \(\overline {DH}=\sqrt {3}*\overline {BH}\)
Using similar triangles IDH and IGF, you get \(\dfrac {2+\overline {BH}} {3.5}=\dfrac {\sqrt {3}\overline {BH}} {\dfrac {\sqrt {3}} {2}}\) = 2\(\overline {BH}\);
2 +\(\overline {BH}\)= 7\(\overline {BH}\); \(\overline {BH}\)= \(\dfrac {1} {3}\)
\(\overline {DH}=\dfrac {\sqrt {3}} {3}\) = the height
 Area of \(\Delta\)DBI = 2 * \(\dfrac {\sqrt {3}} {3}\)*\(\dfrac {1} {2}\) = \(\dfrac {\sqrt {3}} {3}\)
Area of quadrilateral polygon GDBC =  \(\sqrt {3}-\dfrac {\sqrt {3}} {3}\)= \(\dfrac {2\sqrt {3}} {3}\)
The area of both  \(\Delta\)ABC and \(\Delta\)GIC is \(\sqrt {3}\) and both share the quadrilateral polygon GDBC (It's like a Venn diagram).
Thus, the area of both triangles are the same \(\dfrac {\sqrt {3}} {3}\).

Solution II:

There are some harder AMC-10 questions using the same technique.
Using triangles share the same vertex,  you get the two same area "a"s and "b"s because the length of the base is the same.
From the area we've found at solution I, we have two equations:
2a + b = \(\sqrt {3}\)
2b + a = \(\sqrt {3}\)
It's obvious a = b so you can find the area of the quadrilateral being the \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).




Vinjai's solution III:

Draw the two extra lines and you can see that the three lines are medians, which break the largest triangle into 6 equal parts. [You can proof this using triangles share the same vertices : notes later.]
Quadrilateral polygon FECB is  \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).







This is similar to 2005 Mathcounts National Target #4 questions. AMC might have similar ones.
Q: Both triangles are congruent and isosceles right triangles and each has a square embedded in it. If the area of the square on the left is 12 square unites, what is the area of the square on the right? 
Solution I: 

Solution II:

2015 Mathcounts State Prep : Inscribed Cricle Radius and Similar Triangles

Question : \(\Delta\) ABC is an equilateral triangle. Circle "O" is the inscribed circle and it's radius is 15. 

What is the length of the radius of the smaller circle p which is tangent to circle "O" and the two sides?








Here is the link to the basics of inscribed circle radius as well as circumscribed circle radius of an equilateral triangle.

Solution I :
The length of the radius of an inscribed circle of an equilateral triangle is \(\dfrac {1} {3}\) of the height so you know AO is \(\dfrac {2} {3}\) of the height or 30 (the height is 15 + 30 = 45 unit long)

\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {r} {15}=\dfrac {AP} {30}\)
so \(\overline {AP}\) = 2r.
\(\overline {AP}+\overline {PO}=30\)  \(\rightarrow\)2r + r + 15 = 30   \(\rightarrow\)  3r = 15 so r = 5 
or \(\dfrac {1} {3}\) of the larger radius 

Solution II:
\(\Delta\) APE is a 30-60-90 right triangle, so \(\overline {AP}\) = 2r
\(\overline {PO}\) = r + 15
\(\overline {AP}+\overline {PO}\)  \(\rightarrow\)  2r + r + 15 = 30  \(\rightarrow\)  3r  =  15 so  r  = 5 
or \(\dfrac {1} {3}\) of the larger radius 

This is an AMC-10 question.

\(\Delta\) ABC is an isosceles triangle.

The radius of the smaller circle is 1 and the radius of the larger circle is 2,

A: what is the length of \(\overline {AP}\) ?

B. what is the area of \(\Delta\) ABC?







Solution for question A: 
\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {1} {2}=\dfrac {AP} {AP +3}\)
 2\(\overline {AP}\) =  \(\overline {AP}\) + 3 \(\rightarrow\) AP = 3

Using Pythagorean theorem, you can get \(\overline {AE}\) = \(2\sqrt {2}\)
\(\Delta\) AEP is similar to \(\Delta\) ADC [This part is tricky. Make sure you see that !!]
 \(\rightarrow\) \(\dfrac {1} {\overline {DC}}=\dfrac {AE} {AD}\) = \(\dfrac {2\sqrt {2}} {8}\)
\(\overline {DC}\) = \(2\sqrt {2}\)  and \(\overline {BC}\) = 2 * \(2\sqrt {2}\) = \(4\sqrt {2}\)
The area of  \(\Delta\) ABC =  \(\dfrac{1}{2}\)*\(4\sqrt {2}\) * 8 =  \(16\sqrt {2}\)

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

2015 Mathcounts State/National Prep

Harder concepts from Mathcounts Mini :
Also try the follow-up problems with detailed solutions. Don't do every question. Just the ones you think is hard.

Geometry :
#27 : Area and Volume 

#30: Similar Triangles and Proportional Reasoning

#34: Circles and Right Triangles

#35: Using Similarity to Solve Geometry Problems

#41: Analytic Geometry : Center of Rotation

More questions to practice from my blog post. 

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in the of difficulty.

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.

From Mathcounts Mink : Center of rotation, equal distance from a point to other two points or two lines, angle bisector (last problem on the follow-up worksheet)

Mathcounts Mini : #41 - Analytic Geometry

This Week's Work : Week 51 for Inquisitive Young Mathletes

For this week's work :

Try 2013 Mathworks Math Contest problems from Texas State University

Answer key and statistics can be viewed here. 

I'll send you detailed solutions once we finish all the problems.

Practice more "at least" problems :

2014 AMC-10 B problem 16 

Solution

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in order the of difficulty.

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

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 

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.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36





                                                                                               

Solution IV:

If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36




Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!

3 - D Geometry and harder AMCs, AIME links

For Mathcounts advanced group -- harder  Mathcounts State/National problem on 3-D geometry.                      

From Mathcounts Mini :

Using Similarity to Solve Geometry Problems

Area and Volume 

Relationship Between A Box's Dimensions, Volume and Surface Area

AMC, AIME videos from AoPS

Harder/Hardest AMC-10, 12 and AIME problems explained by Mrs. Rusczyk