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Triangle

Let \(\bigtriangleup ABC\) be a triangle and \((\bigtriangleup ABC)=AB+BC+CA\). How is this possible and what's the value of \(AB,BC\) and \(CA?\)

Note by Fatin Farhan
4 years, 2 months ago

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It is possible for right triangles with side lengths: \((5,12,13)\) and \((6,8,10)\).

Actually they are the only right triangles that satisfy this condition (perimeter equals area).

For its proof, see here.

Hope this helps!

EDIT: By the way, I assumed that you were talking about integer side-lengths and area.

Mursalin Habib - 4 years, 2 months ago

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Note that you added an assumption that the triangle is right angled.

How would you deal with this for a general triangle?

The assumption of integer side-lenghts (and hence area) is reasonable. How would your interpretation change if the question was for any real side lengths?

Calvin Lin Staff - 4 years, 2 months ago

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For every set of angles \((A, B, C)\) that sum to \(180^{\circ}\), set \(R = \dfrac{\sin A + \sin B + \sin C}{\sin A \sin B \sin C}\). Then, the triangle with sides \(BC = 2R\sin A\), \(CA = 2R\sin B\), \(AB = 2R\sin C\) has area and perimeter equal to \(\dfrac{2(\sin A + \sin B + \sin C)^2}{\sin A \sin B \sin C}\). This gives us uncountably many triangles.

Jimmy Kariznov - 4 years, 2 months ago

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@Jimmy Kariznov Great approach!

For a more direct approach, if a triangle has area \(A\) and perimeter \(P\), then if you scale the triangle by \( \frac{P}{A} \), it will have area \( A \left( \frac{P}{A} \right)^2 = \frac{P^2}{A} \) and perimeter \( P \frac{P}{A} = \frac{P^2} { A} \).

The \(R\) that you calculated reflects the above scaling.

Calvin Lin Staff - 4 years, 2 months ago

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I know that I added the assumption that the triangle is right-angled. This is because one of my friends wanted me to find all right-triangles with equal area and perimeter (both having an integer numerical value) a couple of weeks ago. So when I saw this question posted here, I immediately replied with the answer I remembered: the right triangles that have equal area and perimeter.

It turns out there are \(5\) triangles with integer side lengths and equal perimeter and area.

They are triangles with side lengths:

\((5, 12, 13)\)

\((6, 8, 10)\)

\((9,10,17)\)

\((7, 15, 20)\)

And \((6, 25, 29)\).

I started off by letting the side-lengths be \(a\), \(b\) and \(c\) and used Heron's formula for the area. And then it turned into a number theory problem.

A good solution to that can be found here and here.

I also learned that such triangles are called perfect triangles.

As for the side lengths being any real (don't you mean positive real?) numbers, Jimmy has shown that there are infinitely many such triangles.

Mursalin Habib - 4 years, 2 months ago

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Here's a problem though: the units are not the same for both sides. It's length squared for area while it's just length for the sum of the sides.

Xuming Liang - 4 years, 2 months ago

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In that case, let \(u\) be a unit distance, and let \(BC = au\), \(CA = bu\), \(AB = cu\). Find all triples \((a,b,c)\) of positive integers such that \([\Delta ABC] = (AB+BC+CA) \cdot u\).

Jimmy Kariznov - 4 years, 2 months ago

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