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

Testimonials for my services. So far, all through words of mouth locally or chance meets online and it's great.

Testimonials from my students/parents  :) 

Dear Mrs Lin,

I just did not get the opportunity earlier to thank you for all your help. _____ was able to make to the National MathCounts largely because of your excellent guidance and coaching. I do not know how to thank you. He had a great once-in-a-lifetime experience there and he really loved the competition as well as meeting other people.

If you can please provide me your mailing address, he wants to send you a gift as a token of thanks for your guidance and tutoring. 

Sincerely,

Dear Mrs. Lin,

I have been meaning to tell you, but just didn't get the time. He just 'LOVES' your sessions. He said he is learning so much and gets to do lots of problems and he likes all the tips/shortcuts you are teaching. He looks forward to your session - he was so upset when we couldn't get back on time from _____ because he didn't want to miss your session. Honestly we came back Tue night only because he cried so much that he didn't want to miss your class:) 

In school, his teacher focuses more on details like, put all the steps, write neatly, don't disturb the class by asking unnecessary questions, don't ask for more work, behave properly etc....so he is not too happy with math in school.

Thank you so much for making such a big difference in his life. He  enjoys doing the homework you are assigning him and has not complained at all. He said "Mrs Lin is so smart and I love her classes...wish she lived next to our house...so I can go to her and have live classes' [disclaimer : I'm not ; my students are much smarter than I and I learn along with them and it's exhilarating ]   Thank you so much!!

One thing he said was it would be helpful if he can have targeted practice worksheets on the tips covered during the class, after that class, so he can practice those shortcuts/tips more.

Hello Mrs. Lin,

This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who

are so selfless in their services to our younger generation.

We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.

BTW, _______ has his chapter level competition tomorrow.

Warm Regards

Mrs. Lin,

Good Afternoon.

Me and my son, ___ and I, have used your blog pages and got a great insight into several of the techniques that you use to solve the problems. I am glad to inform you that _____ was placed 14th in the State Mathcounts for ____. Your blog information has helped us a lot in this preparation and we really want to appreciate all that you do in sharing the information. 

_____ is completing Geometry this year and he hopes to have a exciting next year for AMC 8 and Mathcounts. We hope to learn a lot more from your blog pages.

Regards,

Hi Mrs Lin,

I just finished The One World School House by Salman Khan and I have been thinking of you and the online community you created. Even though _____ and ____ have taken web based courses before, what makes your class different is your inspiration and enthusiasm - you really care about them as unique individuals and you sincerely expect the best from them, more than they (and I sometimes) think they can achieve. Thank you for encouraging them to become a more responsible and self driven learner.

Dear Mrs.Lin,

             I went to NSF tests today and got first place in the Math Bee III. The problems didn't seem that bad although I'm waiting to see my score online. I will most likely be going to nationals which will be held in Ohio this year so that's convenient. Lastly I would like to thank you for all you have done for me. I believe I've grown much more as a student under you and really appreciate everything you have shown me and taught me. I wouldn't be succeeding now if it wasn't for you. :)

Sincerely,

So far,in January state, in algebra 2 he came individual 2nd and one regional February he came 1st (competing with same state high school students while as an 8th grader) 

He could not participate in one regional bec of conflict with MathCounts chapter. We also want to share with you that he got admission into a private school for 9th gr with 100% scholarship. [more than 50 students in that high school are national merit semifinalists, so highly competitive]
Thank you,

An elementary whiz kid, national winners at Math Kangaroo and Math Olympiad. 

_____ will be off to summer sleep away camp where he is not allowed to have any electronics starting _____.  

_________ will be his last class with you for over a month.  He really has been enjoying it and  just so you know I never  have to ask him twice to do the work- crazy because everything else I ask him to do takes at least 5 tries :) 

He will be back online with you at _______ . 

There are many more but I'll take my time to update/upload these infor. sheets. 

To be continued ... 

I'm quite busy these days with resuming our Math Circle + many other projects (my students don't just excel at math, but many other areas equally fun and challenging + most Asian students have much bigger problems with critical reading/not to mention writing, taking initiatives and being strong leaders, and those are my other projects. 

Work-Life balance is utmost important. Less is more. 

Logical games, puzzles and links for summer learning or/and having fun together

You can play daily SET in the New York Times 

Or the official SET website : The Daily SET Puzzle 

   

Math is Fun 

Arithmetic Game 

Sunshine math 

Sunshine math answer key and detailed explanations 

Problem Solving Strategy: Complementary Counting

#: How many three digit numbers contain the digit "9" at least once?
Solution:
For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.

There are 999 -100 + 1 = 900 three digit numbers.

There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 = 252

#2: How many three digit numbers have at least two digits that are the same? 
Solution:
Use complementary counting to find how many three digit numbers have no digit that are the same.
9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648
900 (3 digit numbers) - 648 = 252  

#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?
(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)
Solution: 
From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.

___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.

405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.

600 - 405 =195, which is the answer.

#4: If you toss three coins, what is the probability that at least one coin lands heads up? 
Solution:
There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).

Other applicable problems (answers below):
#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?
#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?
#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.
#4: 9 fair coins are flipped. What is the probability that at least 4 are heads? 
#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?







Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5: 200

Problem Solving Strategies : Complementary Counting

Check out Mathcounts here, the best competition math program for middle school students.

Download this year's Mathcounts handbook here.

Video to watch on complementary counting from "Art of Problem Solving"

Part 1

Part 2

Question: How many two-digit numbers contain at least one 9?

 At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.

9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.
A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1

_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.
So total you have 10 + 9 - 1 (you've counted 99 twice) = 18

This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.

Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.

There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.

Try this question: 
How many three-digit numbers contain at least one 9?
There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers. 
Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers
900 - 8 x 9 x 9 = 252 numbers 

This is called "complementary counting" and there are numerous problems that you can use this strategies to simplify the reasoning. 

Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:
Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.

Solution:  
Use complementary counting. 
If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen. 

Case 1: Team A is not chosen on any of the three days. The probability is (1/3) ³= 1/27.
Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)² times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)

Total possibilities - none - at least 1 time = at least two times Team A will be chosen 
so the answer is 1 - 1/27 - 6/27 = 20/27

Other applicable problems: (answer key below)

#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3? 

#2: What is the probability that when tossing two dice, at least one dice will come up a "3"? 

#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?















Answer key: 

#1:  9000 - 7 x 8 x 8 x 8 = 5416

#2:  The probability of the dice not coming up with a "3" is 5/6.
       1 - (5/6)² = 11/36

2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is 37/49.



 

Games: logical puzzle: the wolf, the sheep and the cabbages

Logical games:

How to cross the river without the wolf eating the sheep, the sheep eating the cabbages.

Here is the link to the puzzle "the wolf, the sheep and the cabbages".












Two versions of "Leap Frog Puzzle".

Leap Frog Puzzle 

Frogs Game 

Rate, Time, and Distance Question

Question: Harder SAT question: Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.

Since rate times time = distance, we can set the equation as 45 t = 30 (1- t), 75t = 30 so
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.

Solution II: 
Let D be the distance from Esther's home to work. 

\(\frac{\Large{D}}{\Large{45}}\) +  \(\frac{\Large{D}}{\Large{30}}\) = 1 (hour)
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90 
D = 18 miles

Solution III:

The rate ratio between driving to work and returning home is 45 : 30 or 3 :  2.

Since rate and time are inversely related (rt = d), the time ratio between the two is 2 : 3. 
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) = 18 miles

Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.

Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)

Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel. 

Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance. 

So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.  










Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed? 
Solution I: 
To find average speed, you use total distance over total time it takes Sally to drive. 
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours 
to drive 84 * 2 = 168 miles. 
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph

Solution II: 
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph


Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time?  How far away is her office ? 
Solution I: 
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up 
the following equation: 
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ;      25 = 20t ;       t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles

To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph

Solution II :
Again, have you noticed that if  both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
 

Some Articles to Read and Ponder after Mathcounts state competition

For those students who really love problem solving, who care and are inquisitive but are disappointed at their chapter's/state's performance, I want to let you know that it's an honor meeting you online and learning from/along with you.

I know no matter how sincere I write here, it won't help much, so I'll just shut up.

However, I'm also disappointed at some students who said they need to take a break for the foreseeable future.

Well, if you love ___ (fill in the blank), you won't stop if you don't get to the chapter/state/nationals.
So there...

Here are some articles that after your taking a break from problem solving (I hope it's not too long), hope to see you come back and read them. Best of luck !!  Keep me posted !!

Pros and Cons of Math Competitions

Dealing with Hard Problems

Life After Mathcounts

Great Mathematicians on Math Competitions and "Genius"

Math Contests Kind of Suck from Mathbabe

TEDxCaltech -Jordan Theriot- The Pleasure of Finding Things Out

Why Physics? Skateboarding Physicist and Educator Dr. Yung Tae Kim

Richard Feyman --The Pleasure of Finding Things Out

Hope it helps !! 

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