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AMC 8 College Mathematics

Cyclic Groups in TIFR Entrance

Concept – Cyclic Groups


Let’s discuss the concept of Cyclic Groups.

A cyclic group G is a group that can be generated by a single element. In particular, if $ G = \{ a, b, c, d, .. \} $, $ * $ is the group operation and $ a $ is a generating element, then if we compute $a $ , $a*a$, , $a*a*a $, etc. we will be able to create all members of the set G.

Get motivated – Problem from TIFR Entrance


Suppose G is a cyclic group with 60 elements. How many generators are there?

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AMC 10 AMC 8 I.S.I. and C.M.I. Entrance Math Olympiad

Division Algorithm

Concept – Division Algorithm


Let’s discuss the concept of Division Algorithm.

Given any integers a and b with ๐‘Žโ‰ 0 there exist unique integers q and r such that ๐‘=๐‘ž๐‘Ž+๐‘Ÿ,๐‘œโ‰ค๐‘Ÿ<|๐‘Ž| . If a does not divide b then r satisfies the stronger inequality 0<๐‘Ÿ<|๐‘Ž|.

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AMC 10 Math Olympiad Videos

Extremal Principle for Counting – AMC 10

Extremal Principle is used in a variety of problems in Math Olympiad. The following problem from AMC 10 is a very nice example of this idea.

AMC 10 Problem 4 (2019)- Based on Extremal Principle

A box contains 28 red balls, 20 green balls, 19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

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Suppose there 5 blue balls, 7 white balls, and 9 green balls. At least how many balls should you pick up (without looking a and without replacement) to be sure that you have picked up at least 4 balls of the same color?

You can also try these problems related to spiral similarity.

You may also click the link to learn its application:- https://www.youtube.com/watch?v=8o8AAWt960o

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AMC 8 USA Math Olympiad

Menalaus Theorem in AMC 8 2019

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Competency in Focus: Menalaus’s Theorem This problem from American Mathematics contest (AMC 8, 2019) will help us to learn more about Menalaus’s Theorem.ย 

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First look at the knowledge graph.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2019/12/AMC-8-2019-Problem-24.png” align=”center” force_fullwidth=”on” _builder_version=”4.0.9″][/et_pb_image][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]In triangle ๐ด๐ต๐ถ, point ๐ท divides side AC so that ๐ด๐ท โˆถ ๐ท๐ถ = 1 โˆถ 2. Let ๐ธ be the midpoint of BD and ๐น be the point of intersection of line BC and line AE. Given that the area of โˆ†๐ด๐ต๐ถ is 360, what is the area of โˆ†๐ธ๐ต๐น?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0.9″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.0.9″ open=”off”]American Mathematical Contest 2019, AMC 8 Problem 25

[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”on” _builder_version=”4.0.9″]Menalaus’s Theorem:ย ย  Given a triangle ABC, and a transversal line that crosses BC, AC, and AB at points D, E, and F respectively, with D, E, and F distinct from A, B, and C, then

$$ \displaystyle {\frac {AF}{FB}\times \frac {BD}{DC}\times \frac {CE}{EA}=-1.}$$

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7/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematicsย 

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Connected Program at Cheenta

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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/amc-8-american-mathematics-competition/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”4.0.9″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” background_layout=”dark” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″][/et_pb_button][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Similar Problems

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AMC 8 Math Olympiad Videos USA Math Olympiad Videos

AMC 8 2019 – Stick and Dot Method

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Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Alice has 24 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0.9″]American Mathematical Contest 2019, AMC 8 Problem 25

[/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0.9″ open=”off”]Combinatoricsย 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0.9″ open=”off”]

7/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematicsย 

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Competency map leading to the problem [/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2019/12/Concept-map-for-AMC-.png” _builder_version=”4.0.9″ hover_enabled=”0″ align=”center” custom_margin=”20px||20px||false|false” background_color=”#474747″][/et_pb_image][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”48px||48px” custom_padding=”20px|20px|0px|20px||” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Watch video

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Connected Program at Cheenta

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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/amc-8-american-mathematics-competition/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”4.0.9″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” background_layout=”dark” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″][/et_pb_button][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Similar Problems

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USA Math Olympiad

Sign Pyramid – AMC 8 2018 Problem 19

Big idea – stabilization

In a counting problem, you may need to stabilize the number of cases. This means, by letting some of the variables change, one may fix the remaining cases.

This is the central idea in the following problem from American Mathematical Contest 8 (AMC 8, 2018, Problem 9).

Also see

Advanced Math olympiad program

Problem

In a sign pyramid a cell gets a “+” if the two cells below it have the same sign, and it gets a “-” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “+” at the top of the pyramid?

amc 8 2018 problem 19

Sequential Hints


(How to use this discussion: Do not read the entire solution at one go. First, read more on the Key Idea, then give the problem a try. Next, look into Step 1 and give it another try and so on.)

Hint 1

Start from the top! Fix the + sign at the top of the pyramid and try filling in the second row from left. Do you see any pattern?

(What happens if you plugin a + sign in the left most block of second row from top?)

Start from the top!

Hint 2

If we fix a + sign in the left most block of second row from top, the other box must contain a + sign. (Why?)

On the other hand fixing a – sign in the left most block of second row from top will force the remaining block to contain – sign.

Thus by varying the left most block of the second row, we can fix the entire row.

Will this work for the third row?

amc 8 2018 problem 19 part 2a
amc 8 2018 problem 19 part 2b

Hint 3

In fact the same trick will work for third row.

Try this. Fix a plus sign in the first row. Then fix second row as well ( plus, plus or minus, minus).

Then try to plugin some sign in the third row’s first block. And check that the remaining two blocks get automatically fixed. This is precisely known as stabilization.

Final Hint

Thus, we should only worry about the first block of second, third and fourth. Fixing them fixes the entire row.

There are 2 choices for the first block of second row, 2 choices for first block of third row and 2 choices for first block of third row. Hence in total we have \( 2 \times 2 \times 2 = 8 \) cases.

Recommended book

Principles and Techniques in Combinatorics Kindle Edition by Chen Chuan-Chong

Try chapter 1.

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USA Math Olympiad

AMC 10 Paper Folding Geometry

We fold a paper using GeoGebra and explore a problem from American Mathematical Contest (AMC 10)

Problem: A rectangular piece of paper whose length is ( \sqrt 3 ) times the width has area A. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B. What is the ratio B: A?

You may also try another paper folding scenario. Here we make the crease a variable!

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USA Math Olympiad

USAMO 2016

Day 1
Problem 1

Let X_1, X_2, \ldots, X_{100} be a sequence of mutually distinct nonempty subsets of a set S. Any two sets X_i and X_{i+1} are disjoint and their union is not the whole set S, that is, X_i\cap X_{i+1}=\emptyset and X_i\cup X_{i+1}\neq S, for all i\in\{1, \ldots, 99\}. Find the smallest possible number of elements in S.

Problem 2

Prove that for any positive integer k, \left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!} is an integer.

Problem 3

Let \triangle ABC be an acute triangle, and let I_B, I_C, and O denote its B-excenter, C-excenter, and circumcenter, respectively. Points E and Y are selected on \overline{AC} such that \angle ABY = \angle CBY and \overline{BE}\perp\overline{AC}. Similarly, points F and Z are selected on \overline{AB} such that \angle ACZ = \angle BCZ and \overline{CF}\perp\overline{AB}.

Lines \overleftrightarrow{I_B F} and \overleftrightarrow{I_C E} meet at P. Prove that \overline{PO} and \overline{YZ} are perpendicular.

Day 2
Problem 4

Find all functions f:\mathbb{R}\rightarrow \mathbb{R} such that for all real numbers x and y, (f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.

Problem 5

An equilateral pentagon AMNPQ is inscribed in triangle ABC such that M\in\overline{AB}, Q\in\overline{AC}, and N, P\in\overline{BC}. Let S be the intersection of \overleftrightarrow{MN} and \overleftrightarrow{PQ}. Denote by \ell the angle bisector of \angle MSQ.

Prove that \overline{OI} is parallel to \ell, where O is the circumcenter of triangle ABC, and I is the incenter of triangle ABC.

Problem 6

Integers n and k are given, with n\ge k\ge 2. You play the following game against an evil wizard.

The wizard has 2n cards; for each i = 1, ..., n, there are two cards labeled i. Initially, the wizard places all cards face down in a row, in unknown order.

You may repeatedly make moves of the following form: you point to any k of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the k chosen cards and turns them back face-down. Then, it is your turn again.

We say this game is \textit{winnable} if there exist some positive integer m and some strategy that is guaranteed to win in at most m moves, no matter how the wizard responds.

For which values of n and k is the game winnable?

The problems on this page are copyrighted by the Mathematical Association of America’s American Mathematics Competitions.

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USA Math Olympiad

Duke Math Meet 2009 : First Relay Round

1A. Find the lowest positive angle \theta that satisfies the equation \sqrt {1+\cos \theta} = \sin \theta + \cos\theta expressed in degrees.

Discussion:

\sqrt {1 +\cos\theta} = \cos\theta + \sin \theta \Rightarrow \sqrt{2\cos^2 \frac{\theta}{2} } = \sqrt2{\frac{1}{\sqrt2} \cos\theta + \frac{1}{\sqrt2} \sin\theta }

Now this gives

\sqrt2 \cos\frac{\theta}{2} = \sqrt2\cos(\theta - \frac{\pi}{4}) \Rightarrow \frac{\theta}{2} = \theta - \frac{\pi}{4} or \frac{\theta}{2} = -\theta + \frac{\pi}{4}

Thus the possible values of \theta are 90^o or 30^o .

Since we require the smallest positive angle hence the answer is 30^o .

1B Let n be two times the tens digit of TNYWR. Find the coefficient of the x^{n-1}y^{n+1} term in the expansion of (2x + \frac{y}{2} + 3)^{2n}

Discussion:

TNYWR is 3. Hence n = 6 Thus we are required to find coefficient of x^5 y^7 term in the expansion of (2x + \frac{y}{2} + 3 )^{12}

This can be easily found from trinomial expansion. The required term is {{12}\choose {5}}(2x)^5 {{7}\choose{7}} (\frac{y}{2})^7 = 792 \times 32 \times \frac{1}{128} = 198

1C Let k be TNYWR, and let n = k/2. Find the smallest integer m greater than n such that 15
divides m and 12 divides the number of positive integer factors of m.

Discussion:

k = 198, hence n = 99.

So we have to look at multiples of 15 greater than 99. We want 12 to divide the number of positive divisors of m.

Suppose m = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k} . The number positive divisors of k is (\alpha_1 +1 )... (\alpha_k + 1)

The first multiple of 15 greater than 99 is 105 = 15 \times 7 . By inspection we see that m = 150.

Categories
USA Math Olympiad

Count of Sparse Subsets – Duke Math Meet 2009 Team Round Problem 6 solution

Call a set S sparse if every pair of distinct elements of S differ by more than 1. Find the number of sparse subsets (possibly empty) of {1, 2, . . . , 10}.

Suppose there are k elements in the sparse subsets. We put them in increasing order. Since any two elements differ by at least 2, we have a_2 \ge a_1 + 2 , a_3 \ge a_2 + 2 \ge a_1 + 4 , ... , a_5 \ge a_1 + 8 Hence there are at most 5 elements in the set.

To count the number of sparse sets we use the following strategy: look at the gaps

for example if there 4 elements in the subsets, after putting them in increasing order, we look at the gaps between the elements. There are 5 such gaps: g_1 = a_1 - 1, g_2 = a_2 - a_1 - 1 ... g_5 = 10 - a_4 . These 5 gaps must account for the remaining 6 elements and the least value of the middle three gaps is 1.

Let g_2' = g_2 + 1 , g_3' =g_3 + 1 , g_4' = g_4+1

Thus we are looking for the number of non negative integer solutions of g_1 + g_2' + g_3' + g_4' + g_5 = 3 . This is given by \binom{7}{3} = 35

Similarly 5 element sparse sets we look at g_1 + g_2' + g_3' + g_4' + g_5' + g_6 = 1 This is given by \binom{6}{1} = 6

Similarly for 3-element subsets we look at g_1 + g_2' + g_3' + g_4 = 5 . This is given by \binom{8}{3} = 56

Similarly for 2-element subsets we look at g_1 + g_2' + g_3 = 7 . This is given by \binom{9}{2} = 36

Finally there are 10 single element sub set and 1 null set all of which are sparse.

Why?

Consider the statement:

‘If there are two elements then they differ by at least 2.’

The above is an implication statement and it is true for singleton or null sets as ‘if there are two elements’ part of the statement is false (an implication statement becomes true if antecedent becomes false).

Hence the total number of sparse sets: 6 + 35 + 56 + 36 + 10 + 1 = 144