# What do you think of this problem ?

• geoffrey159

## Homework Statement

On the picture, compare the area of triangle ABC to the area of A'B'C'.

This problem was shown to me by a 13 years old. Trigonometry forbidden. It seems to me that this is the kind of problem you either solve in 2 minutes, or never solve. In both cases, you don't learn anything. Is this designed to frustrate students or is there any content in this? Your opinion please.

## The Attempt at a Solution

#### Attachments

• triangles.png
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Let the area of triangle ABC be S = (1/2)ab*(sin C) = (1/2)bc*(sin A) = (1/2)ca*(sin B)
By Observation you can prove that
the area of triangle A'B'C' be S' = S + (2S) + (2S) +(2S) = 7S so required answer should be
Trigonometry is Forbidden so you replace sn factors by corresponding heights S'/S = 7
Join B'A, C'B and A'C, You immediately see that the triangle A'B'C' is divided into seven triangular regions of equal area. So that is almost a proof without words!

Yes, this is the way I've done it, and I see no content in this kind of exercise. As you say, it is almost without words!
Furthermore, a 13yo who doesn't know or forgot that the median line splits the area of a triangle in two is done without trigonometry. He will never find, unless someone sees a more elementary proof.

But even that does not require much proof as base is same and heights are same, that is also proof without words. Basic idea is: triangles of different shape can have same area in certain cases.

There may not be wordy content but visualization content is there especially when the students has been told or has understood when two non identical triangles can have same area. Also to foster creativity in imagining the construction which can lead to solution!

So is it an exercise or some kind of riddle for students who have time to loose ? And why ?

Buildimg creativity and intelligence in students require time, It is not like loading a software.