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AMC 10 Geometry Math Olympiad USA Math Olympiad

Surface area of Cube Problem | AMC-10A, 2007 | Problem 21

Try this beautiful problem from Geometry based on Surface area of a cube.

Surface area of cube – AMC-10A, 2007- Problem 21


A sphere is inscribed in a cube that has a surface area of \(24\) square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

  • \(3\)
  • \(4\)
  • \(8\)
  • \(4\)

Key Concepts


Geometry

Cube

square

Check the Answer


Answer: \(8\)

AMC-10A (2007) Problem 21

Pre College Mathematics

Try with Hints


Surface area of Cube - problem

We have to find out the surface area of the inner cube.but to find out the surface area of inner cube the side length is require.but we don’t know the side length of the inner cube.but if you see the above diagram you must notice that  two opposite vertices of the cube are on opposite faces of the larger cube.Therefore two opposite vertices of the cube are on opposite faces of the larger cube.Thus the diagonal of the smaller cube is the side length of the outer square…..

Can you now finish the problem ……….

surface area of inner cube

The area of each face of the outer cube is \(\frac {24}{6} = 4\).Therefore  the edge length of the outer cube is \(2\). so the diagonal of the inner cube is \(2\).let the side length of inner cube is \(x\)

Therefore diagonal =\(\sqrt {x^2 +({\sqrt 2}x)^2}=2\) \(\Rightarrow x=\frac{2}{\sqrt 3}\)

can you finish the problem……..

Therefore surface area of the inner cube is \(6x^2\)=\(6 \times (\frac{2}{\sqrt 3})^2\)=\(8\)

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AMC 8 Math Olympiad USA Math Olympiad

Numbers on cube | AMC-10A, 2007 | Problem 11

Try this beautiful problem from AMC 10A, 2007 based on Numbers on cube.

Numbers on cube – AMC-10A, 2007- Problem 11


The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

  • \(16\)
  • \(18\)
  • \(20\)

Key Concepts


Number system

adition

Cube

Check the Answer


Answer: \(18\)

AMC-10A (2007) Problem 11

Pre College Mathematics

Try with Hints


Given condition is “The numbers from \(1\) to \(8\) are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same”.so we may say that if we think there is a number on the vertex then it will be counted in different faces also.

can you finish the problem……..

Therefore we have to count the numbers \(3\) times so the total sum will be \(3(1+2+….+8)\)=\(108\)

can you finish the problem……..

Now there are \(6\) faces in a Cube…..so the common sum will be \(\frac{108}{6}\)=\(18\)

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AMC 8 Math Olympiad USA Math Olympiad

Octahedron Problem | AMC-10A, 2006 | Problem 24

Try this beautiful problem from Geometry based on Octahedron

Octahedron – AMC-10A, 2006- Problem 24


Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

  • \(\frac{1}{3}\)
  • \(\frac{1}{6}\)
  • \(\frac{3}{4}\)

Key Concepts


Geometry

Octahedron

pyramid

Check the Answer


Answer: \(\frac{1}{6}\)

AMC-10A (2006) Problem 24

Pre College Mathematics

Try with Hints


Octahedron Problem figure

Now we have to find out the volume of the octahedron.given that Centers of adjacent faces of a unit cube are joined to form a regular octahedron.so if we divide the octahedron two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals the we will get two pyramids .Now if we can find out the area of Pyramids them we can find out the volume of Octahedron.can you find out the volume of the Pyramids?

Can you now finish the problem ……….

Prism

Given that side length of a cube is \(1\).Therefore length of all edges of the regular octahedron =\(\frac{\sqrt 2}{2}\).Now the base of the pyramids is a square are will be \((\frac{\sqrt 2}{2})^2\)=\(\frac{1}{2}\).Now clearly the height of the pyramid is half the height of the cube, i.e \(\frac{1}{2}\).So volume of the Pyramid will be \(\frac{1}{3} \times\) (Base area) \(\times\) (height)=\(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2}\)=\(\frac{1}{12}\)

can you finish the problem……..

Therefore the aree of the octahedron= 2 \(\times\) area of Pyramid= 2 \(\times \frac{1}{12}\)=\(\frac{1}{6}\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Total surface area of a cube | AMC-8, 2009 | Problem 25

Try this beautiful problem from Geometry: Total surface area of a cube

Total surface area of a cube – AMC-8, 2009- Problem 25


A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is \(\frac{1}{2}\) foot from the top face. The second cut is  \(\frac{1}{3}\) foot below the first cut, and the third cut is \(\frac{1}{17}\) foot below the second cut. From the top to the bottom the pieces are labeled A,B,C and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?

 a cube
stair
  • $10$
  • $11$
  • $12$

Key Concepts


Geometry

Cube

surface area

Check the Answer


Answer:$11$

AMC-8 (2009) Problem 25

Pre College Mathematics

Try with Hints


Calculate the surface area side by side

Can you now finish the problem ……….

The total height be 1

can you finish the problem……..

total surface of a cube
stair

Clearly The tops of A,B,C, and D in the figure such that 1+1+1+1=4 as do the bottoms .

Thus the total surface area is 8.

 Now, one of the sides has area one, since it combines all of the heights of A,B,C and D which is 1

The other side also satisfies this. Thus the total area now is  10.

Now From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like A, with surface area half. From the back it is the same thing. Thus, the total is \(10+\frac{1}{2}+\frac{1}{2}\)=11

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College Mathematics I.S.I. and C.M.I. Entrance Math Olympiad USA Math Olympiad

Natural Geometry of Natural Numbers

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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.

Number 6 has the shape like a square

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.


The Natural Geometry of 30 – A Cube

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:

Image result for Is four dimensional hyper cube graph

The natural geometry of the number 210 – Tesseract
Image result for tesseract cube
This is the three dimensional projection of the structure of the Tesseract.

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:

Looks like a chair indeed!

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.

Also visit: Thousand Flowers Program

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