Try this beautiful problem from Geometry based on Area of a Triangle Using similarity

Area of Triangle – AMC-8, 2018 – Problem 20

In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.

What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?

Try this beautiful problem from Geometry: Ratio of the area of cube’s cross section . You may use sequential hints to solve the problem.

Area of cube’s cross section – AMC-8, 2018 – Problem 24

In the cube ABCDEFGH with opposite vertices C and E ,J and I are the mid points of segments FB and HD respectively .Let R be the ratio of the area of the cross section EJCI to the area of one of the faces of the cube .what is $R^2$ ?

$\frac{5}{4}$

$\frac{3}{2}$

$\frac{4}{3}$

Key Concepts

Geometry

Area

Pythagorean theorem

Check the Answer

Answer:$\frac{3}{2}$

AMC-8(2018) Problem 24

Pre College Mathematics

Try with Hints

EJCI is a rhombus by symmetry

Can you now finish the problem ……….

Area of rhombus is half product of its diagonals….

can you finish the problem……..

Let Side length of a cube be x.

then by the pythagorean theorem$ EC=X \sqrt {3}$

$JI =X \sqrt {2}$

Now the area of the rhombus is half product of its diagonals

therefore the area of the cross section is $\frac {1}{2} \times (EC \times JI)=\frac{1}{2}(x\sqrt3 \times x\sqrt2)=\frac {x^2\sqrt6}{2}$

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

AMC-8(2017) – Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

$\frac{7}{6}$

$\frac{10}{3}$

$\frac{9}{8}$

Key Concepts

Geometry

congruency

similarity

Check the Answer

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ……….

Now the $\triangle ODB $and $\triangle OCB$ are congruent

can you finish the problem……..

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar….

The length of any side of a triangle is not more than half of its perimeter

Key Concepts

Triangle Inequality

Perimeter

Geometry

Check the Answer

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

A perimeter is a path that encompasses/surrounds a two-dimensional shape. It can be thought of as the length of the outline of a shape.

Try the problem from AMC UP, 2014- Problem 11

These two squares, each with a side length of 10 cm, overlap as shown in the diagram. The shape of the overlap is also a square which has an area of 16 square centimeters. In centimeter s, what is the perimeter of the combined shape?

Australian Mathematics Competition (AMC UP), 2014, Problem No. 11

Perimeter

2 out of 10

Elementary Algebra by Hall and Knigh

Knowledge Graph

Use some hints

Find the Perimeter of 2 boxes

\((4 \times 10) \times 2\)

\(80 cm\)

The perimeter of small box

\(4 \times 4\)

16 cm

So the remaining boxes Space perimeter is 80 – 16 = 64 cm

A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

Try the problem

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

Competency in Focus: 2D Geometry (Areas related to circle)

This problem from American Mathematics Contest 8 (AMC 8, 2017) is based on calculation of areas related to circle. It is Question no. 25 of the AMC 8 2017 Problem series.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/02/amc8_2017_25.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” 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

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ 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 the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $TR$ and $SR$ are each one-sixth of a circle with radius 2. What is the area of the region shown? $\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$[/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=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”on”]American Mathematical Contest 2017, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]

Finding the area of a triangle and sector of a circle. (Area related to circles)

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][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=”4.2.2″]C0nstruction : Let $X$ and $Y$ are the centres of the scetors $ST$ and $TR$ Now Let us join $SX$ and $TY$ What do you think? Will the points $U,S,\textbf{ and}\quad X$ be in a straightline?[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]$U,S,\textbf{ and}\quad X$ will be in a straight line because $\angle STU =60^{\circ}$ And angle of a circle is $360$ i.e., $\angle SXR = \angle TYR = 60^{\circ}$ [Since sector($SXR$)=$\frac{1}{6}circle$] Then $UXY$ will make an equilateral triangle.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]So after construction the figure will look like this : Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$. [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$[/et_pb_tab][et_pb_tab title=”Formulas Used ” _builder_version=”4.2.2″]Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle][/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://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ 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”]

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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 x . 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 . There are rows and columns.

Row – wala Viewpoint

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

Column – wala Viewpoint

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

Now, magically it comes that the . Therefore the number of rows and columns must be equal.

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.

Did you know that there exists a whole set of seven axioms of Origami Geometry just like that of the Euclidean Geometry?

Instead of being very mathematically strict, today we will go through a very elegant result that arises organically from Origami.

Before that, let us travel through some basic terminologies. Be patient for a few more minutes and wait for the gem to arrive.

In case you have forgotten what Origami is, the following pictures will remove the dust from your memories.

Origami (from the Japanese oru, “to fold,” and kami, “paper”) is a traditional Japanese art of folding a sheet of paper, usually square, into a representation of an object such as a bird or flower.

Flat origami refers to configurations that can be pressed flat, say between the pages of a book, without adding any new folds or creases.

When an origami object is unfolded, the resulting diagram of folds or creases on the paper square is called a crease pattern.

We denote mountain folds by unbroken lines and valley folds by dashed lines. A vertex of a crease pattern is a point where two or more folds intersect, and a flat vertex fold is a crease pattern with just one vertex.

In a crease pattern, we see two types of folds, called mountain folds and valley folds.

Now, if you get to play with your hands, you will get to discover a beautiful pattern.

The positive difference between the dotted lines and the full lines is always 2 at a given vertex. The dotted line denotes the mountain fold and the full line denotes the valley fold.

This is encoded in the following theorem.

Maekawa’s Theorem: The difference between the number of mountain folds and the number of valley folds in a flat vertex fold is two.

Isn’t it strange ?

Those who are familiar graph theory may think it is related to the Euler Number.

We will do the proof step by step but you will weave together the steps to understand it yourself. The proof is very easy.

Step 1:

Let n denote the number of folds that meet at the vertex, m of which are mountain folds and v that are valley folds, so that n = m + v. (m for mountain folds and v for valley folds.)

Step 2:

Consider the cross section of a flat vertex.

Step 3:

Consider the creases as shown and fold it accordingly depending on the type of fold – mountain or valley.

Step 4:

Now observe that we get the following cross section. Observe that the number of sides of the formed polygon is n = m + v.

Step 5:

We will also count the angle sum of the n sided polygon in the following way. Consider the polygon formed.

Observe that the vertex 2 is the Valley Fold and other vertices are Mountain Fold. Also the angle subtended the vertex due to valley fold is 360 degrees and that of due to the mountain fold is 0 degrees.

Therefore, the sum of the internal angles is v.360 degrees.

Step 6:

Now we also know that the sum of internal angles formed by n vertices is 180.(n-2), which is = v.360.

Hence we get by replacing n by m+v, that m – v = 2.

QED

So simple and yet so beautiful and magical right?

But it is just the beginning!

There are lot more to discover …

Do you observe any pattern or any different symmetry about the creases or even in other geometry while playing with just paper and folding them?

We will love to hear it from you in the comments.

Do you know Cheenta is bringing out their third issue of the magazine “Reason, Debate and Story” this summer?

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 c 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.

Actually, it shows that to build this shape the requirement of the line segments is as important as the prime numbers to build the number.

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.

I leave some exercises to help you along the way.

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.

The numbers are eagerly waiting for your comments. 🙂

Please mention your observations and ideas and the proofs of the exercises in the comments section. Also think about what type of different shapes can we get from the numbers.