Geometry of Wilson’s Theorem

[et_pb_section fb_built="1" _builder_version="4.0"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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"]

Understand the problem

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Wilson’s Theorem from Number Theory is a beautiful classical result. Enjoy this beautiful geometric proof of the classical result.

We used animations to bring the idea into life.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Now watch the discussion video

[/et_pb_text][et_pb_video src="https://youtu.be/4qbh7mC6YCY" _builder_version="4.1" hover_enabled="0"][/et_pb_video][et_pb_team_member name="Kazi Abu Rousan" position="Cheenta Creative Team" image_url="https://cheenta.com/wp-content/uploads/2020/01/Rousan.jpg" _builder_version="4.1"]Kazi is a student of Physics. He loves the relation between physics and mathematics. He is the creator of this beautiful video.[/et_pb_team_member][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|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Tutorial Problems... try these after watching the video.

[/et_pb_text][et_pb_text _builder_version="4.1" text_font_size="18px" custom_padding="20px|30px|20px|30px|false|false" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset1"]1. Write down a rigorous proof of Wilson’s Theorem. 2. Is the converse of Wilson’s Theorem true? Can you prove it?

You may send solutions to support@cheenta.com. Though we usually look into internal students' work, we will try to give you some feedback.

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Subscribe to Cheenta's youtube channel

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" 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"]

Representation of numbers in base 10 AMC 8 2013 , problem number 13

What are we learning ?

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/AMC-8-2013-problem-13-1.png" align="center" _builder_version="4.1" hover_enabled="0"][/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="4.1" 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" hover_enabled="0" box_shadow_style="preset2"]When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one? [/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="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.1" hover_enabled="0"]American Mathematical Contest 2013, AMC 8 Problem 13 [/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" hover_enabled="0" open="on"]Representation of numbers in base 10 [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" hover_enabled="0" open="off"]4/10[/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 Mathematics

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" header_font="Raleway|300|||||||" background_color="#0c71c3" min_height="12px" custom_margin="30px||30px||false|false" custom_padding="20px|20px|20px|20px|false|false" border_radii="on|5px|5px|5px|5px"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.1" hover_enabled="0"][et_pb_tab title="Hint 0" _builder_version="4.0.9"]Do you really need a hint ? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1" hover_enabled="0"]Let the two digits be $a$ and $b$ that is total score of Clara is ab . [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1" hover_enabled="0"]Now we know that any number in base 10 can be represented as ab=10 a+ b. Given, when Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. So, Clara misinterpreted ab as ba . Again , ba= 10b+a . [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1" hover_enabled="0"]The difference between the two is $|9a-9b|$ which factors into $|9(a-b)|$ . So, what can we say from here? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1" hover_enabled="0"]Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 that is 45. [/et_pb_tab][/et_pb_tabs][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Amc 8 Master class" url="https://cheenta.com/matholympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.0.9" header_font="||||||||" header_text_color="#0c71c3" header_font_size="48px" body_font_size="20px" body_letter_spacing="1px" body_line_height="1.5em" link_option_url="https://cheenta.com/matholympiad/" link_option_url_new_window="on"]

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://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"]

Perfect square numbers AMC 10A, 2014 problem 8

What are we learning ?

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/AMC-10A-2014-problem-8-1.png" align="center" force_fullwidth="on" _builder_version="4.1" min_height="393px" height="198px" max_height="207px" hover_enabled="0"][/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="4.1" 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" hover_enabled="0" box_shadow_style="preset2"]Which of the following numbers is a perfect square? $\textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2$ [/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="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1" hover_enabled="0"]American Mathematical Contest 2014, AMC 10A Problem 8 [/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" hover_enabled="0" open="off"]This number theory problem is based on the concept that when a number is a perfect square [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" hover_enabled="0" open="off"]5/10 [/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 Mathematics

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" 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"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.1" hover_enabled="0"]First of all look at the examples , see that for all positive $n$ , we have $\dfrac{n!(n+1)!}{2}$ .Now what we have to do with this ? [/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.0.9"]Now we have to find which member has what uniform numbers from the given conversation .[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1" hover_enabled="0"]After some simple manipulations , we have $\dfrac{n!(n+1)!}{2}$ $\implies\dfrac{(n!)^2\cdot(n+1)}{2}$ $\implies (n!)^2\cdot\dfrac{n+1}{2}$ . Thus now the problem reduces to finding a value of $n$ such that $(n!)^2\cdot\dfrac{n+1}{2}$ is a perfect square. [/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1" hover_enabled="0"]Since $(n!)^2$ is a perfect square, we must also have $\frac{n+1}{2}$ be a perfect square. In order for $\frac{n+1}{2}$ to be a perfect square, $n+1$ must be twice a perfect square.

[/et_pb_tab][et_pb_tab title="HINT 5" _builder_version="4.1"]

Now check the options and see for what value of n , $n+1$ must be twice a perfect square. $n+1=18$ works, thus, $n=17$ and our desired answer is $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$ [/et_pb_tab][/et_pb_tabs][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Inversion and Ptolemy's Theorem

Understand the problem

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Ptolemy's Theorem gives a relation between product of diagonals of a quadrilateral with the sum of the product of its opposite sides. However it can be proved using tools from inversive geometry (that Ptolemy probably did not know). Try this!

Now watch the discussion video

[/et_pb_text][et_pb_video src="https://www.youtube.com/watch?v=pw-lXWZt5wg" _builder_version="4.1"][/et_pb_video][et_pb_team_member name="Kazi Abu Rousan" position="Cheenta Creative Team" image_url="https://cheenta.com/wp-content/uploads/2020/01/Rousan.jpg" _builder_version="4.1"]Kazi is a student of Physics. He loves the relation between physics and mathematics. He is the creator of this beautiful video.[/et_pb_team_member][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|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Tutorial Problems... try these after watching the video.

[/et_pb_text][et_pb_text _builder_version="4.1" text_font_size="18px" custom_padding="20px|30px|20px|30px|false|false" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset1"]1. How can you generalize Ptolemy's theorem if the quadrilateral is not cyclic?? 2. Suppose two points A, B, are inside a unit circle G. Invert A and B about G to A' and B' respectively. If the length of AB is 0.5, what is the length of A'B'? You may send solutions to support@cheenta.com. Though we usually look into internal students' work, we will try to give you some feedback. [/et_pb_text][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Subscribe to Cheenta's youtube channel

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Mean and Median calculation AMC 8, 2013 Problem 5

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What are we learning ?

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Competency in Focus: Mean and Median calculation

This problem from American Mathematics Contest 8 (AMC 8, 2013) is based on calculation of mean and median. It is Question no. 5 of the AMC 8 2013 Problem series.

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="10px|10px|10px|10px|false|false" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica" custom_margin="10px||10px||false|false"]

First look at the knowledge graph:-

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/AMC-8-2013-Problem-5-1.png" alt="calculation of mean and median- AMC 8 2013 Problem" title_text=" mean and median- AMC 8 2013 Problem" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="429px" height="189px" max_height="198px" hover_enabled="0" custom_padding="10px|10px|10px|10px|false|false"][/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" inline_fonts="Aclonica"]

Next understand the problem

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Basic Statistics and Data Representation mainly calculation of mean and median.

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="4.0.9"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Start with hints

Let us first find the median of the weight of the five children. For this, we first have to arrange the weights of the five children in increasing order. As we know, the median is the middle value, if there is an odd number of observations, and if there is an even number of observations, it is the average of the two middle values. Thus, lining up the numbers (5, 5, 6, 8, 106), we see that it is 6 pounds.

[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.1"]Now what we have to find is the mean of the weights of five children .The average weight of the five kids is $\dfrac{5+5+6+8+106}{5} = \dfrac{130}{5} = 26$ .[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1"]The median here is obviously less than the mean.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1"]Therefore, the average weight is bigger than median weight , by $26-6 = 20$ pounds, making the answer , average by 20.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

AMC - AIME - USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

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A trigonometric polynomial ( INMO 2020 Problem 2)

The problem

Suppose $P(x)$ is a polynomial with real coefficients satisfying the condition
$$
P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta),
$$
for every real $\theta$. Prove that $P(x)$ can be expressed in the form
$$
P(x)=a_0+a_1\left(1-x^2\right)^2+a_2\left(1-x^2\right)^4+\cdots+a_n\left(1-x^2\right)^{2 n},
$$
for some real numbers $a_0, a_1, a_2, \ldots, a_n$ and nonnegative integer (n).

Hint 1

Using a very standard trigometric identity, we can easily convert the following ,
$$
\begin{aligned}
P(\cos \theta+\sin \theta) & =P(\cos \theta-\sin \theta
\Longrightarrow P\left(\sqrt{2} \sin \left(\frac{\pi}{4}+\theta\right)\right) & =P\left(\sqrt{2} \cos \left(\frac{\pi}{4}+\theta\right)\right) \
\Longrightarrow P(\sqrt{2} \sin x) & =P(\sqrt{2} \cos x)
\end{aligned}
$$
⟹ $P(\sqrt{2} \sin x)=P(\sqrt{2} \cos x) \quad$

Assuming,

$\left(\frac{\pi}{4}+\theta\right)=x$ for all reals $x$. So,

$P(-\sqrt{2} \sin (x))=P(\sqrt{2} \sin (-x))=P(\sqrt{2} \cos (-x))=P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ for all $x \in \mathbb{R}$. Since $P(x)=P(-x)$ holds for infinitely many $x$, it must hold for all $x$ (since $P(x)$ is a polynomial). so we get that, $P(x)$ is a even polynomial.

Hint 2

$P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ implies that
$$
P(t)=P\left(\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right) \text { putting }, x=\cos ^{-1}(t / \sqrt{2})\right.
$$
for infinitely many $t \in[-\sqrt{2}, \sqrt{2}]$.
$$
\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right)=\sqrt{2-t^2} \text { so we get, } P(x)=P\left(\sqrt{2-t^2}\right)
$$

Again as it is a polynomial function we can extend it all $\mathbb{R}$. And we get, $P(x)=P\left(\sqrt{2-x^2}\right)$ for all reals (x)

Hint 3

Since $P(x)$ is even, we can choose an even polynomial $Q(x)$ such that, $Q(x)=P(\sqrt{x+1}) \cdot P(\sqrt{1+x}$=$Q(x)=a_0+a_1 x^2+a_2 x^4+\cdots+a_n x^{2 n}$ now take, $\sqrt{1+x}=y$ and you get the polynomial of required form.

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Kites in Geometry | INMO 2020 Problem 1

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Let $ \Gamma_1 $ and $ \Gamma_2 $ be two circles with unequal radii, with centers $ O_1 $ and $ O_2 $ respectively, in the plane intersecting in two distinct points A and B. Assume that the center of each of the circles $ \Gamma_1 $ and $ \Gamma_2 $ are outside each other. The tangent to $ \Gamma_ 1 $ at B intersects $ \Gamma_2 $ again at C, different from B; the tangent to $ \Gamma_2 $ at B intersects $ \Gamma_1 $ again in D different from B. The bisectors of $ \angle DAB $ and $ \angle CAB $ meet $ \Gamma_1 $ and $ \Gamma_2 $ again in X and Y, respectively. different from A. Let P and Q be the circumcenters of the triangles ACD and XAY, respectively. Prove that PQ is perpendicular bisector of the line segment $ O_1 O_2 $.

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1. Suppose $ P O_1 Q O_2 $ be a kite (that is $ PO_1 = PO_2 $ and $ QO_1 1 = QO_2 $. Show that PQ is perpendicular bisector of the other diagonal $ O_1 O_2 $.$.

2. Show that for any two circles intersecting each other at two distinct points, the common chord is bisected perpendicularly by the line joining the center.

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INMO 2020 Problem 4

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Let $latex n\ge 3$ be an integer and $latex a_1,a_2,\cdots a_n$ be real numbers satisfying $latex 1<a_2\le a_2\le a_3\cdots \le a_n$. If $latex \Sigma_ia_i=2n$ then prove that $latex 2+a_1+a_1a_2+a_1a_2a_3+\cdots +a_1a_2\cdots a_{n-1}\le a_1a_2\cdots a_n$.

[/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" admin_label="Blog" _builder_version="3.22" custom_margin="|||" custom_padding="0px|0px|21px|0px|false|false"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_padding="0|0px|24px|0px|false|false" use_custom_width="on" custom_width_px="960px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Hint 1" _builder_version="3.12.2"]

The conditions hint at inequalities involving an order, such as the rearrangement and Chebychev inequalities. Also note that $latex a_i=2$ for all $latex i$ is an equality case, hence we should try to use inequalities in such a way that matches the equality case.

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The RHS can be rewritten as $latex a_1a_2\cdots a_n= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+a_1$. That is, $latex a_1a_2\cdots a_n-1= a_1a_2\cdots a_n-a_1a_2\cdots a_{n-1}+a_1a_2\cdots a_{n-1}-a_1a_2\cdots a_{n-2}+a_1a_2\cdots a_{n-2}\cdots -a_1+a_1=a_1a_2\cdots a_{n-1}(a_n-1)+a_1a_2\cdots a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1)$.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.12.2"]

Now Chebychev inequality gives $latex \frac{a_1a_2\cdots a_n-1}{n}= \frac{a_1a_2\cdot a_{n-1}(a_n-1)+a_1a_2\cdot a_{n-2}(a_{n-1}-1)+\cdots+ a_1(a_2-1)+(a_1-1)}{n}\ge \frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}\cdot\frac{(a_1-1+a_2-1+\cdots +a_n-1)}{n}=\frac{1+a_1+a_1a_2+\cdots +a_1a_2\cdots a_{n-1}}{n}$. Cancelling the denominators, we get the desired result.

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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.

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Geometry of circles and rectangles AMC 8 2014 problem 20

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Rectangle $ABCD$ has sides $CD=3$ and $DA=5$ . A circle of radius $1$ is centered at $A$ , a circle of radius $2$ is centered at $B$ , and a circle of radius $3$ is centered at $C$ . Which of the following is closest to the area of the region inside the rectangle but outside all three circles? $[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy]$ [/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="4.1" 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="on"]American Mathematical Contest 2014, AMC 8 Problem 20

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Geometry of circles and rectangles [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]6/10[/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 Mathematics

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[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"]The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]Here the area of the rectangle is 3.5=15. Area of quater circles is (Area of the circle )/4 = $ \frac{\pi . r^2}{4} $ , where r= radius of the circle . so, The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$ , where area of the quater for circle A is $ \frac{\pi}{4} $ ,for circle B is $ \frac {\pi .2^2}{4} $ , for circle C is $ \frac{\pi.3^2}{4} $.Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$ .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]Now what can we do with $15-\frac{7\pi}{2}$ to get an approximate value ?[/et_pb_tab][et_pb_tab title="Hint 4 " _builder_version="4.1"]As we know that we can approximate $ \pi $ by $ \frac{22}{7} $ . and substituting that in will give 15-11=4.[/et_pb_tab][/et_pb_tabs][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" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Competency in Focus: Mean and Median calculation

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Basic Statistics and Data Representation mainly calculation of mean and median.

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

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