Disappearing Triangles Problem

Geometry Level 2

One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints, in turn, are joined to form still another triangle. This process continues indefinitely.

Find the sum of the perimeters of all these triangles that are defined above.

140

144

145

147\sqrt{3}

6 solutions

Prasun Biswas
Feb 10, 2014

First, we must understand that this is a case of infinite geometric series which states that if $a$ be first term and $r$ be the common ratio between $2$ consecutive terms of an infinite GP, then we have the sum of such series as ----

$S=\frac{a}{1-r}$ if $|r| \lt 1$

In this case, we see that the side of the biggest equilateral triangle is $24$ cm. Now, since the midpoints of the sides of the triangle are taken to form another triangle and then midpoints of that inner triangle are joined to form yet another triangle and so on, thus by mid-point theorem, we can see that the triangle inside the outermost triangle has side length $=\frac{24}{2}=12$ cm and the triangle inside it will have side length $=\frac{12}{2}=6$ cm and so on. Also, since the outermost triangle is an equilateral triangle, so if we use the mid point theorem to find each side of the inscribed triangles, then we see that all the triangles are equilateral triangles. So, here we see that side length of each successive inscribed triangle is half of the previous one, so $r=\frac{1}{2}$ and $|r|=|\frac{1}{2}|=\frac{1}{2} \lt 1$ . Also, starting side length is the first term $(a)=24$

Now, Perimeter of each equilateral triangle $=3\times (\text{Side length of the triangle})$

So, total sum of perimeters of all triangles

$=(3\times 24)+(3\times 12)+(3\times 6)+.......\infty$

$=3(24+12+6+....\infty)=3(\frac{24}{1- \frac{1}{2}})=3(\frac{24}{\frac{1}{2}})=3(24\times 2)=3\times 48 = \boxed{144\text{ cm}}$

Great Question Prasun! For a moment I thought this must be Geometry! :D

I used a similar approach:

1st triangle has side = 72

2nd triangle has side = 72/2 = 36

3rd triangle has side = 36/2 = 18

and so on.

This forms the infinite G.P. : 72, 36, 18, 9, ... Here, $a=72$ and $r= \frac{1}{2}$

Hence, total permieter = $\frac{a}{1-r}= \frac{72}{1- \frac{1}{2} } = \frac{72}{ \frac{1}{2} } = \boxed{144}$

Kou$htav Chakrabarty - 7 years, 3 months ago

Thats the simplest and the easiest!!

Mani Vel - 5 years, 9 months ago

yeah! this is right ans..

sanjeev mishra - 7 years, 1 month ago

The total sum can be written as: S=72+36+18+9+9/2+9/4+9/8...... we then take the first four numbers in the sequence 72,36,18,19 and add them up to a total of 135 but we still need to add the missing numbers (9/2+9/4+9/8....) to do so we notice that we can factor out the nine. S=135+9(1/2+1/4+1/8.....) it is easy to show that the sequence in the brackets is convergent to one, therefore the sum can now be expressed as follows: S=135+9(1)= 144 Hence, the answer is 144 cm.

Francisco González Isás - 7 years, 1 month ago

You just did the same thing which I did in my solution but in a rather lengthy way.

If a GP is such that the first term is $a$ and the common ratio is $r$ such that $|r|<1$ , then the sum of the GP, i.e., the sum $S=a+ar+ar^2+....+ar^n (n\to \infty)$ converges to $S=\frac{a}{1-r}$ .

Now, take the series of the perimeters : $72,36,18,....\infty$ .

We can see that the value of each successive term is half of the previous one, so $r=\frac{1}{2}$ and so we see that $|r|=|\frac{1}{2}|=\frac{1}{2}<1$

The sum of the series, $S=\frac{72}{1-\frac{1}{2}}=\frac{72}{\frac{1}{2}}=72\times 2 = \boxed{144}$

Prasun Biswas - 7 years, 1 month ago

Hey would you mind telling how do we get S=a/1-r ??

Muhammad Ali Yousha - 7 years, 3 months ago

If we consider an infinite GP with first term and common ratio as $a,r$ respectively such that $|r|<1$ , i.e, $-1\lt r\lt 1$ , then the sum $S_n$ of $n$ terms of the GP will be given by ---

$S_n=\frac{a(1-r^n)}{(1-r)} \implies S_n=(\frac{a}{1-r}-\frac{ar^n}{1-r})$

Since, $|r|<1$ , so as $n$ increases then $r^n$ decreases, so $n\to \infty \implies r^n\to 0$

Then, the sum of the infinite GP is -----

$S=\displaystyle \lim_{n\to \infty}(\frac{a}{1-r}-\frac{ar^n}{1-r})$

$=\frac{a}{1-r}-\displaystyle \lim_{n\to \infty}(\frac{ar^n}{1-r})$

$=\frac{a}{1-r}-\frac{a\times 0}{1-r}=\frac{a}{1-r}-0 = \boxed{\frac{a}{1-r}}$

Prasun Biswas - 7 years, 3 months ago

For a undegraduate, the easist demonstration actually is:

We will call "n" as the perimeter of the first triangle. (On this case is 72 -> 24 + 24 + 24) If we check the second, we see that the perimeter is exatly a half from the first. (Each side is actually cuting the middle of the first triangle and going to another middle, creating a triangle with simple a half of the perimeter of the first) So the third will be a half from the second, with exactly the same idea.

With this, the sum will be:

S = n + n/2 + n/4 + n/8 + ... (equation I)

Let's multiply I by 1/2

S/2 = n/2 + n/4 + n/8 + ... (equation II)

On equation II, we have all the terms that I has except by "n". So if we do I - II we will have:

S - S/2 = n + (n/2 + n/4 + n/8 + ... ) - (n/2 + n/4 + n/8 + ... ) S - S/2 = n (Equation III)

Simplifing...

S (1 - 1/2) = n S/2 = n

\boxed{S = 2n}

So for this question we will have: S = 2*72 \boxed{144}

It's exactly the same as a/1-r , but the ratio is not explicit on the formula (ratio is 1/2)

Felipe Magalhães - 7 years, 3 months ago

Mardokay Mosazghi
Mar 6, 2014

Since this is geometric sequence the perimeter of the first triangle is 72 every time the sides are divide by two so to find the perimeter to infinity would be (72/.5) which would equal 144.

Abhijeet Srivastava
Feb 11, 2014

let side = a Total perimeter = 3a + 3a/2 + 3a/4 + 3a/8 + ............till infinity. If you take 1/2 = r, Then the expression becomes, 3a + 3ar + 3ar^2 + 3ar^3 + 3ar^4 + .......................... till infinity this is an Geometric Progression (GP) and it can be solved by using the formula, a/(1 - r) where 'a' is the constant term i.e 3a and 'r' is the common ratio i.e 1/2. By solving we get, T. perimeter = 6a = 6 * 24 = 144 cm

Jason Tang
Feb 11, 2014

Notice that each subsequent triangle's sidelength is half of the next biggest one. This means that the perimeter is $3(24)+3(24)(\frac{1}{2})+3(24)(\frac{1}{4})+\cdots+3(24)(\frac{1}{\infty})=3(24)(1+\frac{1}{2}+\frac{1}{4}+ \cdots+\frac{1}{\infty})$ . In the parentheses is the infinite geometric sum with first element 1 and r value of $\frac{1}{2}$ . The formula for the sum of an infinite geometric series is $\dfrac{a}{1-r}$ , which in this case would be $\dfrac{1}{1-\frac{1}{2}}=2$ . Therefore, the sum of the perimeters is $3\times 24\times 2=\boxed{144}$

Lucas Maekawa
Jan 30, 2018

Let $S = 72+36+18+9+\ldots$ .

If we multiply both sides by $2$ , we get $2S=144+72+36+18+9+\ldots$ . Replacing $S$ , we get $2S = 144 + S\Rightarrow 2S-S=144\Rightarrow S = 144$ .

Samik Bharwad
Mar 25, 2016

The perimeter of the biggest triangle = 72 cm . The second biggest triangle = 36 cm and so on .... Now let us take 72 as the common factor . So the equation become 72(1+1/2+1/4+1/8........) . Now we can group the 1/2+1/4+1/8 ...... Portion to get 1 so it becomes 72(1+1) =72*2=144cm

Disappearing Triangles Problem

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