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: Some Counting and Probability Questions on Dot Grids

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

#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.

#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?

Solution:

#5 National Target: There are 16C4 = \(\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}\)= 1820 ways to select 4 points on the geoboard.

There are 3 x 3 = 9  one by one squares and 2 x 2 = 4 two by two squares and 1 x 1 = 1 three by three squares. (Do you see the pattern?)

                                                         

There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of √ 5.

9 + 4 + 1 + 4 + 2 = 20 and \(\dfrac {20} {1820}=\dfrac {1} {91}\)

#5: Solution:
AMC-10A: There are 9C3 = \(\dfrac {9\times 8\times 7} {3\times 2\times 1}\)= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so 

\(\dfrac {8} {84}=\dfrac {2} {21}\)

#29: Solution: 
Use the length of the two congruent legs to solve this problem systematically. 

 There are 16   1 - 1 - \(\sqrt {2}\)    isosceles triangles.
There are 8    \(\sqrt {2}\)  by  \(\sqrt {2}\) by 2 isosceles triangles. (See that ?)

There are 4     2 - 2 - \(2\sqrt {2}\)  isosceles triangles.
There are 4    \(\sqrt {5}\)  by  \(\sqrt {5}\) by 2 isosceles triangles.
Finally, there are 4  \(\sqrt {5}\)  by  \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.  

How Many Zeros?

Problems: (Solutions below.)

#1. 2003 Chapter Team # 7--How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

#2. How many zeros are at the end of 2013!? 

#3. How many zeros are at the end of 10! *9!*8!*7!*6!*5!*4!*3!*2!*1!*0!?

#4. What is the unit digit of 10! + 9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1! + 0!?

#5. The number \(3^{4}\times 4^{5}\times 5^{6}\) written out in full. How many zeros are there are at the end of the number?

#6. How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)

#7. 2009 National Sprint #18-- What is the largest integer n such that \(3^{n}\) is a factor of 1×3×5×…×97×99?

Solutions:

#1.For 100!, there are -- 100/5 = 20 , 20/5 = 4 (Stop when the quotient is not divisible by 5 and then add up all the quotients.), or 20 + 4 = 24 zeros.
For 200!, there are 200/5 =40 , 40/5 = 8, and  8/5 = 1, or total 40 + 8 + 1 = 49 zeros.
For 300!, there are 300/5 = 60, 60/5 = 12, and 12/5 = 2, or total 60 + 12 + 2 = 74 zeros.
Add all the quotients and you get 147 zeros. 

#2. Use the same method as #1 and the answer is 501 zeros.

#3:Starting at 5!, you have one "0", the same goes with 6!, 7!, 8!, and 9!
10! will give you 2 extra "0"s. Thus total 7 zeros.

#4: Since starting with 5! you have "0" for the unit digit, you only need to check 4! + 3! + 2! + 1! + 0!.
24 + 6 + 2 + 1 + 1 = 34 so the unit digit is 4.

#5: Make sure to prime factorize all the given number sequences, in this case, it's \(3^{4}\times 2^{10}\times 5^{6}\) after you do that.
2 * 5 = 10 will give you a zero since there are fewer 5s than 2s so the answer is 6 zeros.

#6: You need the same number of 2 and 5 multiple together to get a "0".
31! gives you 30/5 = 6, 6/5 = 1 or 6 + 1 = 7 multiples of 5
31! gives you 10//2 = 15...15/2 = 7...7/2 = 3...3/2 = 1 or 15 + 7 + 3 + 1 = 26 multiples of 2. 16!*8!*4!*2!*1! gives you 4 multiples of 5 and 8 + 4 + 2 + 1 (16!) + 4 + 2 + 1 (8!) + 2 + 1 (4!) + 1(2!)
= 26 multiples of 2.
Thus the multiples of 2s all cancel out, the answer is "0" zeros. 

#7: There are 3*1, 3*3, 3*5...3*33 or \(\dfrac {33-1} {2}+1=17\) multiples of 3.
There are 9*1, 9* 3...9*11 or \(\dfrac {11-1} {2}+1 = 6\) extra multiples of 3.
There are 27*1, 27*3 or 2 extra multiples of 3.
There is 81*1 or 1 extra multiples of 3.
Add them up and the answer is 26.

2014 AMC 8 Results , Problems, Answer key and Detailed Solutions.

E-mail me if you want to join my group lessons for Mathcounts/AMCs prep.
Different groups based on your current level of proficiency.

My students are amazing. They NEVER stop learning and they don't just do math.

Please don't contact me if you just want state/national questions since I'm extremely busy these days with the preps.
You can purchase Mathcounts National tests and other prep materials at Mathcounts store.

You can also use my online blog contents. If you really understand those concepts, I'm sure you can be placed in your state's top 10 in the above average states [Every year I'd received some online students'/parents' e-mails about their (their kids') state results], not the most competitive states, which are crapshoot for even the USAJMO qualifiers, that I know.

Take care and best of luck.

Have fun problem solving and good luck.

2014 AMC-8 problems and solutions from AoPS wiki

Comments :

#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.

 #10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points.  (such as #12, 13, 18 -- if you were not trolled and others)

 #11 : Similar questions appear at AMC-10, Mathcounts.  For harder cases, complementary counting is easier.
This one, block walk is easier.

Counting Path on a Grid 

This one includes forbidden path finder

 #12: 1/ 3!

How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches,  only2 baby photos match, 3 baby photos match, or none matches ?

 #13: number theory

For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd

For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd

For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.

SAT/ACT has similar type of problems.

#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep 

#17: rate, time and distance could be tricky

Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.

Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.

 Check out the notes from my blog and see for yourself.

#18 : Trolled question. Oh dear !! 

1 4 6 4 1     , but it doesn't specify gender number(s) so 
(4 + 4)/2^4 is the most likely. 

#19 : more interesting painted cube problems --> one cube is completely hidden inside.

Painted cube animation from Fairy Math Tutors

Painted cube review   Use Lego or other plops to help you visualize how it's done.


#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.

#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.

#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z 

For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.

#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.

#24 : a more original question --

To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.

#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap

I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.

Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.

Happy Holiday !!

ACT worksheets

ACT worksheets

Let me know if you have any other questions. Best and take good care, Mrs. Lin 

Lesson   Equations of Circles

Practice : Practice with Equations of Circles

worksheet : 

http://www.jmap.org/StaticFiles/PDFFILES/WorksheetsByPI/Geometry/Coordinate_Geometry/Drills/PR_G.G.71.pdf

Similar triangles : 

http://www.regentsprep.org/regents/math/geometry/gp11/Lstrategy.htm

http://www.regentsprep.org/regents/math/geometry/gp11/PracSim.htm

worksheet : 

https://www.jmap.org/JMAP/RegentsExamsandQuestions/PDF/WorksheetsByPI-Topic/Geometry/Informal_and_Formal_Proofs/G.G.46.SideSplitterTheorem.pdf