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What are we learning ?

[/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″]Competency in Focus: Area of triangles This problem from American Mathematics contest (AMC 10A, 2013) is based on calculation of area of triangles . [/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″ 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″]Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$ $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ [/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 2013, AMC 10A  Problem 3 [/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.1″ hover_enabled=”0″ open=”off”]Area of triangles [/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_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″]Given Square $ABCD$ has side length $10$.  So,  $AB=10$. [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.1″ hover_enabled=”0″]Now, we know area of a triangle =$\frac{(height).(base)}{2}$. Try to use this here . [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.1″ hover_enabled=”0″]So , we have the area of $\triangle ABE$ is equal to $\frac{AB(BE)}{2}$Plugging in $AB=10$ , what we get ? [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.1″ hover_enabled=”0″]we get $80 = 10BE$. Dividing, we find that $BE=\boxed{\textbf{(E) }8}$ [/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″]

Similar Problems

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Does there exist a Magic Rectangle?

Magic Squares are infamous; so famous that even the number of letters on its Wikipedia Page is more than that of Mathematics itself. People hardly talk about Magic Rectangles.

Ya, Magic Rectangles! Have you heard of it? No, right? Not me either!

So, I set off to discover the math behind it.

Does there exist a Magic Rectangle?

First, we have to write the condition mathematically.

Take a table of dimension $m$ x $n$. Now fill in the tables with positive integers so that the sum of the rows, columns, and diagonals are equal. Does there exist such a rectangle?

Let’s start building it from scratch.

Now let’s check something else. Let’s calculate the sum of the elements of the table in two different ways.

Let’s say the column, row and diagonal sum be $S$. There are $m$ rows and $n$ columns.

Row – wala Viewpoint

The Rows say the sum of the elements of the table is $S.m$. See the picture below.

Column – wala Viewpoint

The Rows say the sum of the elements of the table is $S.n$. See the picture below.

Now, magically it comes that the $S.m = S.n$. Therefore the number of rows and columns must be equal.

Whoa! That was cute!

Visit this post to know about Magic Square more.

Edit 1: Look into the comments for a nice observation that if we allowed integers, and the common sum is 0, then we may not have got the result. Also we need to define the sum of the entries of a diagonal of a rectangle.

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Natural Geometry of Natural Numbers

How does this sound?

The numbers 18 and 30 together looks like a chair.

The Natural Geometry of Natural Numbers is something that is never advertised, rarely talked about. Just feel how they feel!

Let’s revise some ideas and concepts to understand the natural numbers more deeply.

We know by Unique Prime Factorization Theorem that every natural number can be uniquely represented by the product of primes.

So, a natural number is entirely known by the primes and their powers dividing it.

Also if you think carefully the entire information of a natural number is also entirely contained in the set of all of its divisors as every natural number has a unique set of divisors apart from itself.

We will discover the geometry of a natural number by adding lines between these divisors to form some shape and we call that the natural geometry corresponding to the number.

Let’s start discovering by playing a game.

Take a natural number n and all its divisors including itself.

Consider two divisors a < b of n. Now draw a line segment between a and b based on the following rules:

• a divides b.
• There is no divisor of n, such that a < c < b and a divides c and c divides b.

Also write the number $\frac{b}{a}$ over the line segment joining a and b.

Let’s draw for number 6.

Now, whatever shape we get, we call it the natural geometry of that particular number. Here we call that 6 has a natural geometry of a square or a rectangle. I prefer to call it a square because we all love symmetry.

What about all the numbers? Isn’t interesting to know the geometry of all the natural numbers?

Let’s draw for some other number say 30.

Observe this carefully, 30 has a complicated structure if seen in two dimensions but its natural geometrical structure is actually like a cube right?

The red numbers denote the divisors and the black numbers denote the numbers to be written on the line segment.

Beautiful right!

Have you observed something interesting?

• The numbers on the line segments are always primes.

Exercise: Prove from the rules of the game that the numbers on the line segment always correspond to prime numbers.

Did you observe this?

• In the pictures above, the parallel lines have the same prime number on it.

Exercise: Prove that the numbers corresponding to the parallel lines always have the same prime number on it.

Actually each prime number corresponds to a different direction. If you draw it perpendicularly we get the natural geometry of the number.

Let’s observe the geometry of other numbers.

Try to draw the geometry of the number 210. It will look like the following:

Obviously, this is not the natural geometry as shown. But neither we can visualize it. The number 210 lies in four dimensions. If you try to discover this structure, you will find that it has four different directions corresponding to four different primes dividing it. Also, you will see that it is actually a four-dimensional cube, which is called a tesseract. What you see above is a two dimensional projection of the tesseract, we call it a graph.

A person acquainted with graph theory can understand that the graph of a number is always k- regular where k is the number of primes dividing the number.

Now it’s time for you to discover more about the geometry of all the numbers.

Exercise: Show that the natural geometry of $p^k$ is a long straight line consisting of k small straight lines, where p is a prime number and k is a natural number.

Exercise: Show that all the numbers of the form $p.q$ where p and q are two distinct prime numbers always have the natural geometry of a square.

Exercise: Show that all the numbers of the form $p.q.r$ where p, q and r are three distinct prime numbers always have the natural geometry of a cube.

Research Exercise: Find the natural geometry of the numbers of the form $p^2.q$ where p and q are two distinct prime numbers. Also, try to generalize and predict the geometry of $p^k.q$ where k is any natural number.

Research Exercise: Find the natural geometry of $p^a.q^b.r^c$ where p,
q, and r are three distinct prime numbers and a,b and c are natural numbers.

Let’s end with the discussion with the geometry of {18, 30}. First let us define what I mean by it.

We define the natural geometry of two natural numbers quite naturally as a natural extension from that of a single number.

Take two natural numbers a and b. Consider the divisors of both a and b and follow the rules of the game on the set of divisors of both a and b. The shape that we get is called the natural geometry of {a, b}.

You can try it yourself and find out that the natural geometry of {18, 30} looks like the following:

Sit on this chair, grab a cup of coffee and set off to discover.