Harder Geometry Question from National Mathcounts

Question: Similar to a 96 National Mathcounts question: Circles of the same colors are congruent and tangent to each other. What is the ratio of the area of the largest circle to the area of the smallest circle?

Solution :
The most difficult part might be to find the leg that is "2R - r".
R and r being the radius of the median and the smallest circle.

Using Pythagorean theorem, you have \(\left( R+r\right) ^{2}= R^{2}+\left( 2R-r\right) ^{2}\)
\(R^{2}+2Rr+r^{2}=R^{2}+4R^{2}-4Rr+r^{2}\)
Consolidate and you have 6Rr = \(4R^{2}\)\(\rightarrow\) 3r = 2R \(\rightarrow\) r =\(\dfrac {2R} {3}\) The ratio of the area of the largest circle to the area of the smallest circle is thus \(\rightarrow\)\(\dfrac {\left( 2R\right) ^{2}} {\left( \dfrac {2R} {3}\right) ^{2}}\) = 9 .

2013 Mathcounts State Prep: Counting and Probability

Question #1: Rolling two dice, what is the probability that the product is a multiple of 3?
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6  Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).

Solution II: 
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{5}}{\Large{9}}\)

Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{1}}{\Large{2}}\)

2013 Mathcounts State Prep: Similar Triangles and Height to the Hypotenuse

There are many concepts you can learn from this image, which cover numerous similar right triangles, ratio/proportion/dimensional change and the height to the hypotenuse.

Question:
#1: Δ ABC is a 3-4-5 right triangle. What is the height to the hypotenuse? 
Solution: 
Use the area of a triangle to get the height to the hypotenuse. 
Let the height to the hypotenuse be "h"
The area of Δ ABC is  \(\Large\frac{3*4}{2}\)= \(\Large\frac{5*h}{2}\)
Both sides times 2 and consolidate: h =  \(\Large\frac{3*4}{5}\) = \(\Large\frac{12}{5}\)

Practice: What is the height to the hypotenuse?

Question:
#2: How many similar triangles can you spot?
Solution: 
There are 4 and most students have difficulty comparing the largest one with the other smaller ones.
Δ ABC is similar to Δ ADE, Δ FBD, ΔGEC. Make sure you really understand this and can apply this to numerous similar triangle questions. 

Question: 
#3: What is the area of  □ DEGF if \(\overline{BF}\) = 9 and \(\overline{GC}\) = 4
Solution: 
Using the two similar triangles Δ FBD and  ΔGEC (I found using symbols to find the corresponding legs
 to be much easier than using the lines.), you have \(\frac{\Large{\overline{BF}}}{\Large{\overline{FD}}}\) = \(\frac{\Large{\overline{GE}}}{\Large{\overline{GC}}}\).
s (side length of the square) = \({\overline{GE}}\) =  \({\overline{FD}}\)
Plug in the given and you have 9 * 4 = s² so the area of □ DEGF is 36 square units. (each side then is square root of 36 or 6)

Question: 
#4: Δ ABC is a 9-12-15 right triangle. What is the side length of the square? 
Solution :
The height to the hypotenuse is\(\frac{\Large{9*12}}{\Large{15}}\) = \(\frac{\Large{36}}{\Large{5}}\)
Δ ABC is similar to Δ ADE. Using base and height similarities, you have \(\frac{\Large{\overline{BC}}}{\Large{\overline{DE}}}\) = \(\frac{\Large{15}}{\Large{S}}\) = \(\frac{\frac{\Large{36}}{\Large{5}}}{\frac{\Large{36}}{\Large{5}} - \Large{S}}\)
Cross multiply and you have 108 - 15*S = \(\frac{\Large{36}}{\Large{5}}\) *S
S =\(\frac{\Large{180}}{\Large{37}}\)

Ratio, Proportions -- Beginning Problem Soving (SAT level)

Questions:
#1: At a county fair, if adults to kids ratio is 2 to 3 and there are 250 people at the fair. How many adults and how many kids? 

Solution I :  Adults to Kids = 2 : 3 (given). Let there be 2x adults and 3x kids (only when two numbers have the same multiples can you simplify the two number to relatively prime.
 2x + 3x = 250;  5x = 250; x = 50 so there are 2x or 2 * 50 = 100 adults and 3x or 3 * 50 = 150 kids.

Solution II:  From the given information, you know that every time when there are 5 parts (2 + 3 = 5), there
will be 2 parts for A, or   of the total, and 3 parts for B, or of the total.
  * 250 = 100 adults and * 250 = 150 kids

#2: In a mixture of peanuts and cashews, the ratio by weight of peanuts to cashews is 5 to 2. How many pounds of cashews will there be in 4 pounds of this mixture? (an actual SAT question)

Solution:
You can use either the first or second method (see above two solutions) but the second solution is much faster. 
* 4 = pounds

#3: Continue with question #1: How many more kids than adults go to the fair? 

Solution I: 
Use the method on #1 and then find the difference.
150 - 100 = 50 more kids.
  
Solution II: 
Using the method for #1: solution II, you know    of the all the people go to the fair are adults ,
and of the total people go to the fair are kids.
(  -   ) * 250 = 50 more kids.

#3: If the girls to boys ratio at an elementary school is 3 to 4 and there are 123 girls, how many boys are there at that elementary school? 
Solution:
This one is easy, you set up the equation. Just make sure the numbers line up nicely.
Let there be "x" boys. = . Cross multiply to get x. Or since 123 is 41 * 3;
4 * 41 = 164 boys

#4 : If the girls to boys ratio at an elementary school is 2 to 5 and there are 78 more boys than girls, how many girls are there at the elementary school? 

Solution I: 
Again, you can use the algebra and let there be 2x girls and 5x boys. 
According to the given, 5x - 2x = 3x = 78 so x = 26
Plug in and you get there are 2*26 = 52 girls and 5*26 = 130 boys.

Solution II: 
If you keep expanding the ratio using the same multiples, 2 : 5 = 4 : 10 : 6 : 15...
Do you see that the difference of boys and girls are always multiples of 3 ( 5 - 2  = 3).
  = 26 so there are 2 * 26 or 52 girls

Review: Shoestring method, Heron's formula, Pythagorean Triples and Others related to Area of a triangle or polygon

Please give me feedback if there is any error or what concepts to be included for future practice tests. Thanks a lot in advance.

Online practices on finding the area of a triangle or irregular polygons.
Click here for the timed online test. Type in your first name and only write down "number" answers.

Before trying out the online timed test, here are some reminders.

a. If the irregular polygon include the origin, then using that as a starting point for shoestring method might save some calculation time.

b. There are Pythagorean Triples hidden in lots of questions that ask for the area of a triangle given three side lengths. Thus, if you find the side length that matches the triples or multiples of the triples, it might be much easier to use that instead of heron's formula.

c. For quadrilaterals, if the diagonals are perpendicular to each other, such as rhombus, kite or others, it's much easier to use D1* D2 / 2 to get the area.

d. Sometimes breaking the polygon into different smaller parts is the easiest way to find the answer.

e. Deliberate practices are the key to steady progress.

Have fun problem solving!!

Line Intercept, Slope, Area/Geometry Questions

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This question is similar to 2010 #3 Mathcouts Chapter Team problem. 

Solution I: Using the given slope, you can find the equation for AB ,
which is: y = -2 x + b, plug in (4, 7) and you have 7 = -2 x 4 + b so b = 15
The y intercept for AB is (0, 15)
Now you have four points, (0, 0), (0, 15), (4, 7) and (15, 0).
Using shoestring method to find area of any polygon, you have the area of polygon
AEDO = 82.5 square units.

Solution II:

After finding the y intercept for AB.

You break the area into three parts.

The side of the shaded region is 4 x 7 = 28.

The area of triangle ACE = 4 * (15-7) / 2 = 16

The area of triangle
EBD = (external height)7 * (15-4) / 2 = 38.5

So the total area is 82.5 square units.




Applicable problems: Answer key below.

#1:  The slope of AB is -3 and it intercepts with CD at point E, which is (3, 3). If the x intercepts of CD is (10,0), what is the area of AEDO?

#2: The slope of AB is 2 and it intercepts with CD at point E, which is (-2, 2 ). If the x intercepts of CD is (-7,0), what is the area of AEDO?

#3:  The slope of AB is -3 and it intercepts with CD at point E, which is ( 5, 8 ). If the x intercepts of CD is (14,0), what is the area of AEDO? -- Andrew's question.

#4: The slope of AB is \(\dfrac {-3} {2}\)and it intercepts with CD at point E, which is ( 2, 7 ). If the x intercepts of CD is (16,0), what is the area of AEDO? --Daniel's question.

#5: The slope of AB is -4/5 and it intercepts with CD at point E, which is (5, 2 ). If the x intercepts of CD is (12,0), what is the area of AEDO?












Answer key: 
#1: 33
#2: 13
#3: 113.5
#4: 66
#5: 27

Trianges That Share the Same Vertex/Similar Triangles

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Besides similar triangles, "triangles that share the same vertex" also appears regularly on geometry questions. 

Look , for example, at the left image. 
It's easy to see AD being the height to base CE for Δ AEC.

However, it's much harder to see the same AD being the external height of  Δ ABC if BC is the base.





Question to ponder based on the above image:
#1: If  BC : CE = 2 to 5 and the area of Δ ABE is 98, what is the area of Δ ABC and Δ ACE?

Solution:
Since both triangles share the same height AD, if you use BC and CE  as the bases, the area ratio stays constant as 2 to 5.  Thus the area of Δ ABC = (2/7) * 98 = 28 and the area of
Δ AEC = (5/7) *98 = 70.

Knowing the above concept would help you solve the ostensibly hard trapezoid question.

Question #2: If ABCD is a trapezoid where AB is parallel to CD,  AB = 12 units and CD = 18 units.
If the area of  Δ AEB = 60 square units, what is the area of ΔCED, Δ AED and ΔBEC?






Solution:  Δ AEB and ΔCED are similar (Why? Make sure you understand this?) Thus if AB to CD = 12 to 18 or 2 to 3 ratio (given), the ratio of the area of the two similar triangles is 2² to 3² = 4 to 9 ratio.
The area of  Δ AEB = 60 square units (given), so the area of ΔCED is 9 * (60/4) = 135 square units. 
Δ AEB and Δ AED share the same vertex. The height is the same with base BE and ED.
Thus the area ratio is still 2 to 3. The area of Δ AEB = 60 square units (given),
 the area of Δ AED = 3 * (60/2) = 90 square units.

Using the same reasoning you get the area of ΔBEC = 90 square units. [Keep in mind that Δ AED and ΔBEC don't have to be congruent, but they do have the same area.]

Similar Triangles I

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There are numerous questions on similar triangles at competition math.
Practice how to spot them and use them to solve tougher chapter, median hard state questions.

For example, There are three similar triangles in this image.

Triangle ABC is similar to triangle BDC and triangle ADB.

Using consistent symbols will help you set up the right proportion comparisons much easier.

BD² = AD x DC
BA² = AD x AC
BC² = CD x AC  

All because of the similarities.
That is also where those ratios work.











#1: What is the height to the hypotenuse of a 3-4-5 right triangle? 
Solution: 
 The area of a triangle is base x height over 2.
 Area of a 3-4-5 right triangle  = (3 x 4) / 2 = (5 x height to the hypotenuse) / 2
 Both sides times 2 : (3 x 4) = 5 x height to the hypotenuse
 The height to the hypotenuse = 12/5 

The same goes with 6-8-10 right triangle, the height to the hypotenuse is 4.8.
It's very straightforward so make sure you know how and why that works.

Hope this helps.

2007 State Harder Algebra Questions

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#16 2007 Mathcounts State Sprint: A data set for a class of 25 sixth graders has their ages listed as
the integer values of either 10 or 11 years. The median age in the data set is 0.36 years greater than the mean. How many 10-year-olds are in the class?

#23: 2007 Mathcounts State Sprint:  A box contains some green marbles and exactly four red marbles.
The probability of selecting a red marble is x%. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x - 15)%. How many green marbles are in the box before the number of green marbles is doubled?












#16: Solution I: 
 Since there are 25 students, the median has to be either 10 or 11 (Why?)
Since the mean has to be 10 (if all the children are all 10 year old) or higher, the median has to be 11.
Let there be x 10 year old students, and there are (25 - x) 11 year old students.

both sides * ----

 

 

Solution II: 

Let there be x 10 year old and y 11 year old. 

According to the given: 

10x + 11 y = 266
For x to be integer, y has to either 6 or 16. Further checking, only when y = 16, x = 9 works. 

#23: Let there be g numbers of Green marbles.
According to the given, we can set up two equations:

                                          equation 1

           equation 2

Equation 1 minus equation 2 and you have: 0 = xg + 60 -2xg + 30g
Move all the variables to the other side:
xg - 30g = g (x - 30) = 60
If g = 4, x = 45 (doesn't work)
If g = 6, x = 40 (it works.)

Mental Math Practices on Ratio, Percents and Proportion

Mental Math Practices: All questions can be solved in your head reasonably fast (less than 5 minutes) if you understand the concepts well.

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1. What is 12.5% of 128?

2. After 20% discount, you paid $144? What is the original price?

3. To make 20% profit, you charge your customer 144 dollars. What is the wholesale price?

4. You lost 20% in the stock market. What percentage increase will you break even?

5. A : B = 2 : 5 and A is 75 more than B, what is B?

6. A : B = 2 : 3 and A + B = 14, what is A and what is B?

7. A : B = 3 : 5 and B : C = 2 to 9, what is A : B?

8. Two triangles are similar and the length of their base ratio is 1: 2. The smaller triangle has a base that equals 5 and an area that is 110. What is the area of the larger triangle?

9. One side of the square increases 10% and the other side decreases 20 percent. What is the percentage change? Increase or decrease? By how much?

10. There are 60 students in a gym and 30% are boys. After a few more boys enter the gym, now 40% are boys. How many boys enter the room?

Beginning Algebra: II

Learn the basics here.

Here we are going to review and work on some other harder problems.

Question #1:   Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

Solution: 
Let there be x tickets sold for singles and y tickets sold for couples. According to the given:
20 x + 35 y = 2280 --- equation 1
x  + 2 y = 128 --- equation 2

equation 2 times 20 - equation 1 and you have 20 x + 40 y - (20 x + 35 y) = 20 x 128 - 2280 = 280
5y = 280 so y = 56 and plug in to equation 2 and you get x = 16

There will be 16 tickets sold for singles and 56 tickets sold for couples.
Always check your answer to see if that is right.

Question #2: There are 20 questions on an algebra test, you got 5 points for each question you answer correctly and - 2 points for each questions you answer wrong. 
If a student gets 79 for his score and he answers all the questions, how many questions does he answer right? 

Solution I: 
If the student gets all the questions right, he'll get 5 x 20 = 100 points; however, he only gets 79 points. 
For each questions he gets wrong, he not only doesn't have the 5 points but on top of that, he gets 2 points deducted, thus it's minus 7 total. 

(100 - 79) / 7 = 3 so the student gets 3 wrong and 17 questions right. 
Check if that's correct: 17 x 5 - 2 x 3 = 79

Solution II:  
Let the student get x questions right, and that means he gets (20 - x) questions wrong.
5x - 2 (20 -x) = 79;    5x - 40 + 2x = 79;   7 x = 119 so x = 17

Question #3: My piggy bank contains only nickels, dimes, and quarters, and contains 20 coins worth a total of $3.30. If the total value of the quarters is five times the total value of all the other coins, how many dimes are in the piggy bank?

Solution:  
According to the given: 
N + D + Q = 20  --- equation 1
The total value of the quarters is 5 times the total value of all the other coins, which mean that if the total value of N + D = x, the total value of Q is 5 x
x + 5x = 330 (Make sure to use the same unit) so x = 55 (cents)
5x = 5 x 55 = 275 and 275/ 25 = 11 so there are 11 quarters

Plug in equation 1 and you have N + D = 20 - 11 = 9
Since 5N + 10D = x = 55, D and N could be
D        N
5         1
4         3
3         5
2         7... etc   2 + 7 = 9 so there are 2 dimes (7 nickels and 11 quarters)


Other applicable problems: (answer key below)
#1: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number of nickels, then how many of each does he have?

#2:  Kim has $1.65 in nickels, dimes and quarters. She has 10 coins all together and the number of quarters is equal to the number of nickels and dimes combined. Haw many of each coin does she have?

#3:  A collection of 24 nickels, dimes, and quarters is worth $3.20. There are seven more nickels than dimes. How many of each are there in the collection?

#4: Continue with #3, if there are seven more dimes than nickels, how many of each are there in the collection? 











Answer key: 
#1: 6 quarters, 20 nickels and 15 dimes  
#2: 2 nickels, 3 dimes and 5 quarters 
#3: 9 quarters, 4 dimes and 11 nickels
#4: 7 quarters, 12 dimes and 5 nickels

Geometry : Dimensional Target type questions

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Questions: (Solutions below)

#1: A circle is circumscribed around a square and another circle is inscribed in the same square. Find the ratio of the area of the smaller circle to the area of the larger circle. Express your answer as a common fraction.

#2: A square pyramid has a base edge of 32 inches and an altitude of 1 foot. A square pyramid whose altitude is one-fourth of the original altitude is cut from the vertex. The volume of the remaining frustrum is what fractional part of the volume of the original pyramid?

#3: This question is similar to one SAT question that most of my high school students have problem with. 

  























Solutions:
The two questions above are very similar. They are both related to dimension change, and appear very often on competition math. Just remember that for any two figures, the ratio of their areas is simply (ratio of base * ratio of height), and the ratio of their volumes is (ratio of base area * ratio of height). In cases where the ratio of all the sides are the same - that is, when you're dealing with similar figures - the ratio of the areas is just (ratio of side)²2, whereas the ratio of the volumes is just (ratio of side)³.

#1: The ratio of the two radii is 1 to √ 2   (45-45-90 degree right triangle), so the area ratio is 
(1 over √ 2 )² = 1/2.

#2 : Since the two pyramids are similar, and you know that the height of the new smaller pyramid is 1/4 of the old height, the volume must be (1/4)³ = 1/64 of the original pyramid.
So the volume of the frustum is (1-1/64) = 63/64 of the original shape.

#3: Since the radius is a linear relationship. R1 to r2 = 1 : 2 (given), the area ratio is 1 : 4 (square both ratio)The larger measure is double the smaller one, so once you know the area of ABC, you can get the area of DEF by multiplying the area of ABC by 4 x 2 = 8 so 5 x 8 = 40 square units.

Similar Triangles II

=

 
 
=



=

Weird but Delicious Math Questions

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Problem: (Solution below)
#1: 1993 Mathcounts National Team Round #4 :The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

#2: 2000 AMC10 #22: One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Solution: 

#1: For Anna to know that Brett and Chris have different numbers, she must have an odd number
because 14 - odd = odd and you can only get odd sum of two numbers if they are different, one odd
one even.

For Brett to say that he already know all three have different numbers, he not only must have an odd number, but also his number has to be larger or equal to 7 and is not the same as what A have.

Otherwise, A-B-C could be 1-1-12; 3-3-8; or 5-5-4.
It would exceed 14 if you have 7-7-__. (All numbers are positive) so Brett"s and Anna's numbers must
be different.

If Brett has 7, then the numbers could be A-B-C = 1-7-6 ; 3-7-4 or 5-7-2.

If Brett has 9, then the numbers could be A-B-C = 1-9-4; or 3-9-2.

If Brett has 11, then then numbers could be A-B-C = 1-11-2.

From the above possibilities you know Chris has to have 6 for him to be sure he knows all the numbers. 

So A-B-C = 1-7-6 and the product of ABC = 1 x 7 x 6 = 42

#2: Let there be m cups of mild, c cups of coffee. and x people in Angela's family.
According to the given, you can set up the following equation: 
  m + c = = 8. Two ways to solve this equation.

Solution I:

m + c =   Both sides times 12 x

Both m and c need to be positive so the only x that works is when x = 5.

Solution II: 

m + c  = 8, Both sides times 12 and you have 

3m + 2c = 96; m + c is a multiple of 8.

30      3

28      6

26      9

.

.

.

2       45;  from  (30 + 3 ) = 33 to (2 +45) = 47 only 40 is a multiple of 8

and 40 divided by 8 = 5 so 5 is the answer.