2016 Mathcounts State Prep : How to Avoid Making Careless Mistakes

Why Can Some Kids Handle Pressure While Others Fall Apart? from the New York Times

Relax! You'll Be More Productive from the New York Times

First, please read Mathcounts "Forms of Answers" carefully for a few times and remind yourself not to write down the unallowable answers.

Stop Making Stupid Mistakes by Richard Rusczyk from Art of Problem Solving

How to Avoid Careless Mathematical Errors from Reddit

Common errors from all my students: 

Number 1 issue with most of the boys/and a few girls who have strong intuition in math is their horrible handwriting since their minds work much faster than their hands can handle. (See!! I'm making excuses for them.)

I call some of my students "second try ___ [Put your name here.] or third try ____ because I don't need to teach them how to solve a problem but the normal pattern is that it has to take them twice or three times to get it finally right. Thus, they need to slow down and organize their thoughts.

Practice your numbers:  4 (it's not 21), 7, 9 (don't dislocate the circle on top), 2 (don't make it looks like a 7), etc...  

35 cents is very different from 0.35 cents.

Unless stated, you write improper fraction as your answer. Make sure to simplify the answer.

Many mistakes happen when it involves fractions, negative numbers, parenthesis, and questions that
have radicals on the denominator and you need to simplify it, so make sure to double check your math.

Some problems takes many steps to reach the final solutions. Some involve lots of data, extreme long strings of information so if those are your weakness, skip them first and go back to them later.

Every point is the same so don't stay with a tedious question too long and panic later not able to finish the last few harder questions which might be much easier to solve arithmetic wise if you know how.

Circle questions trouble many students. Make sure you are aware of what's being tested.
Is it circumference or area, is it diameter or radius, is it linear to area (squared) or vice versa ?
Are the units the same (most frequently appeared errors)?

Don't forget to divide by 2 for the triangle, times 6 if you use area of an equilateral triangle to get the hexagon. For geometry questions or unit conversions, don't do busy work, write down the equivalent equations and cancel like crazy.

Same goes to probability questions, cancel, cancel and cancel those numbers. You don't need to practice mental math multiplication for most of the Mathcounts problems. Think Smart!!

The answer for probability questions can only be 0 to 1, inclusive.

For lots of algebra questions, manipulations are the way to go or number sense. Use the digit clue to help you narrow down the "trial and error" method. 

Make sure you have the terms and space type questions right. Calendar questions, how many numbers (inclusive, exclusive, between, ...), Stage doesn't necessarily start with 1 or 0.
Make sure you don't over-count or under-count.

Mos students got counting questions wrong when it involves limit.

Case in point : How many non-congruent triangles are there if the perimeter is 15? 
Answer: There are 7 of them.  Try it !!

To be continue...

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

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:

Triangular Numbers & Word Problems: Chapter Level

Triangular Numbers :  From Math is Fun. 1, 3, 6, 10, 15, 21, 28, 36, 45...

Interesting Triangular Number Patterns: From  Nrich

Another pattern: The sum of two consecutive triangular numbers is a square number.

What are triangular numbers? Let's exam the first 4 triangular numbers:

The 1st number is  "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is \(\frac{n(n+1)}{2}\)

It's the same as finding out the sum of the first "n" natural numbers.

Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)

On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree

On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.

On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree

On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.


The question is "How many gifts were given out on the Day of Christmas?"

Solution I"
1st Day:       1
2nd Day:      1 + 2
3rd Day       1 + 2 + 3
4th Day        1 + 2 + 3 + 4
.
.
12th Day      1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12

Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364

Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use \(\frac{n(n+1)(n+2)}{6}\)

n = 12 and \(\frac{12 (13)(14)}{6}= 364 \)

Here is a proof without words.

Applicable questions: (Answers and solutions below)

#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?

#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?

#3 What is the 20th triangular number?

#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords? 

#5: Following the pattern, how many triangles are there in the 15th image? 

Answers: 
#1 1262  The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).

#2 55; 220    a. \(\frac{10*11}{2}=55\)   b. \(\frac{10*11*12}{6}=220\)

#3 210   \(\frac{20*21}{2}=210\)  

#4 1276 
1 chord :    2 regions or \(\boxed{1}\) + 1
2 chords:   4 regions or  \(\boxed{1}\)  + 1 + 2
3 chords:   7 regions or \(\boxed{1}\)  + 1 + 2 + 3
.
.
50 chords:\(\boxed{1}\)  + 1 + 2 + 3 + ...+ 50 = 1 +  \(\frac{50*51}{2}\) =1276

#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total. 
It follows the triangular number pattern. The 15th triangular is  \(\frac{15*16}{2}\) =120 

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