BrainDen.com - Brain Teasers
• 0

# "Big" Triangles

Go to solution Solved by k-man,

## Question

Let us call "big" a triangle with all sides longer than 1. Given a equilateral triangle with all the sides equal to 5.

Prove that:

a) You can cut 100 big triangles out of the given one.

b) You can divide the given triangle onto 100 big nonintersecting ones fully covering the initial one.

c) The same as b), but the triangles either do not have common points, or have one common side, or one common vertex.

## Recommended Posts

• 0
• Solution

Not sure if this qualifies for c), but should be sufficient for a) and b)

Orient the equilateral triangle, so that one of its sides is horizontal - we'll call it base. Start from the top of the triangle and measure just over 2 along both sides and draw a horizontal line segment connecting the points. This new line segment parallel to the base will be just over 2 units long. Find the midpoint of this line segment and connect it with both base corners of the equilateral triangle. We just formed 4 "big" triangles.

Next, copy the horizontal line segment into the bottom big triangle and repeat the steps 33 times. The picture below shows the process repeated only 5 times and the final 3 triangles formed by connecting the horizontal line segment with the midpoint of the base. It's easy to see that all triangles formed using this method are big and cover the original triangle entirely. Not sure if they meet the condition of c) though.

##### Share on other sites
• 0

The smallest big triangle is equilateral with side = 1+.

Its area is therefore 1/25 the area of an equilateral triangle with side = 5.

Part b) says nonintersecting and fully covering. This means the areas are equal.

That is, the initial triangle can be dissected into the 100 big triangles.

This seems impossible.

##### Share on other sites
• 0

The smallest big triangle is equilateral with side = 1+.

Its area is therefore 1/25 the area of an equilateral triangle with side = 5.

Part b) says nonintersecting and fully covering. This means the areas are equal.

That is, the initial triangle can be dissected into the 100 big triangles.

This seems impossible.

Are you sure about that, bonanova?

What happens when a triangle gets reeeeeaaaalllly flat?

##### Share on other sites
• 0

BMAD, © confuses me. How do you divide an area into figures that don't share sides or vertices?

For example, how can you divide a square into several polygons that don't share any points, sides, or vertices but fully cover the square?

##### Share on other sites
• 0

Are you sure about that, bonanova?

What happens when a triangle gets reeeeeaaaalllly flat?

Not sure what you mean by flat. What am I missing?

##### Share on other sites
• 0

Are you sure about that, bonanova?

What happens when a triangle gets reeeeeaaaalllly flat?

Not sure what you mean by flat. What am I missing?

Consider a triangle with side lengths 1+, 1+, and 2+. This triangle has a height of 0+, and its area is arbitrarily small. In other words, it is a very flat triangle. However, it is also a big triangle, and it is quite easy to see then that 100 (or more) of these triangles can be cut from our side-5 equilateral triangle.

##### Share on other sites
• 0

BMAD, © confuses me. How do you divide an area into figures that don't share sides or vertices?

For example, how can you divide a square into several polygons that don't share any points, sides, or vertices but fully cover the square?

Forgive my grammar here. What I mean is an exclusive "and". They cannot share any common points but may share only one common side and/or one common vertex. Edited by BMAD
##### Share on other sites
• 0

Not sure if this qualifies for c), but should be sufficient for a) and b)

Orient the equilateral triangle, so that one of its sides is horizontal - we'll call it base. Start from the top of the triangle and measure just over 2 along both sides and draw a horizontal line segment connecting the points. This new line segment parallel to the base will be just over 2 units long. Find the midpoint of this line segment and connect it with both base corners of the equilateral triangle. We just formed 4 "big" triangles.

Next, copy the horizontal line segment into the bottom big triangle and repeat the steps 33 times. The picture below shows the process repeated only 5 times and the final 3 triangles formed by connecting the horizontal line segment with the midpoint of the base. It's easy to see that all triangles formed using this method are big and cover the original triangle entirely. Not sure if they meet the condition of c) though.

big triangles.png

That was really impressive.

Good job!

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.