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  1. The incenter of triangle \(ABC\) is \(I\) and inradius is \(2\). What is the smallest possible value of \(AI+BI+CI\) ?
  2. A triangle has base of length \(8\) and area \(12\). What is the radius of the largest circle that can be inscribed in this triangle?
  3. The least consecutive ten numbers, all greater than \(10\), are determined that are respectively divisible by the numbers 1 through 10. Write down the smallest number among these \(10\).
  4. In trapezium \(ABCD, AD||BC, AD < BC\), unparallel sides are equal. A circle with centre O is inscribed in the trapezium. \(OAD\) is equilateral. Find the radius of the circle if the area of the trapezium is \(\frac{800}{\sqrt{3}}\)

Note by Tahmid Kenshin
3 years, 8 months ago

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I reshared this note because I thought that these problems are quite interesting and I wanted them to get a better exposure. But no one's commenting on these. So, here I go. I worked on 2 in between my classes today.

Solution to \(2\):

You can immediately see that the height of the triangle is \(3\). You should be familiar with the relation, \(rs=[ABC]\) where \(r\) and \(s\) are the inradius and semi perimeter of \(\triangle ABC\) and \([ABC]\) denotes its area.

So, in order to maximize \(r\), we need to minimize the perimeter.

Now draw two parallel lines \(l_1\) and \(l_2\) such that they are \(3\) units away from each other. Set points \(A\) and \(B\) on \(l_1\) such that \(AB=8\). Now notice that any point \(C\) on \(l_2\) satisfies the criteria of a triangle with base \(8\) and area \(12\).

Our job is to minimize \(AC+BC\).

How do we do that?

If you know calculus, you can put in some co-ordinates for points \(A\), \(B\) and \(C\) and differentiate \(AC+BC\) and set it to zero. I'm going to take a purely geometrical approach.

Let \(A'\) be the reflection of \(A\) over \(l_2\). Now \(AC+BC=A'C+BC\) and \(A'C + BC\) is minimized whenever \(C\) is on the line \(A'B\) [triangle inequality]. Now if \(C\) is on \(A'B\), \(C\) is the midpoint of \(A'B\). So, \(A'C=BC\) and that means \(AC\) has to be equal to \(BC\) if you want to minimize\(AC+BC\) . I think you can take it from here. I'm skipping ahead of the calculation. The answer is \(\frac{4}{3}\).

That's all I could do in between my classes. I'm sorry if my solution's too short. Normally, I'd post a picture and explain the steps more thoroughly but I'm a bit busy. I apologize again. I will work on another one of them tomorrow in between classes.

I haven't looked at the fourth problem properly but I see it's about a tangential trapezium [Br] and tangential trapeziums have some interesting properties. Take a look at them. They might be of use. Mursalin Habib · 3 years, 8 months ago

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@Mursalin Habib I'm sorry that I'm a little late. Got caught up with something.

Solution to \(1\):

I see that you got some help from others on this one. I'm going to present a solution that doesn't resort to Jensen's inequality or Trigonometry.

Problems 1 and 2 are closely related. If you've understood everything I wrote on problem 2, this is going to be very simple. Now if you join\(A, I\); \(B, I\) and \(C, I\), you're going to get three triangles with height \(r=2\) and the sides of \(\triangle ABC\) as their bases.

Notice that I proved above that \(AI+BI\) is minimized if and only if they're equal. [draw a line parallel to \(AB\) through \(I\) and let \(I'\) be the reflection of \(I\) over that line ... see above for more details].

So, \(AI+BI\) is minimized if and only if \(AI=BI\).

\(BI+CI\) is minimized if and only if \(BI=CI\).

And, \(CI+AI\) is minimized if and only if \(CI=AI\).

That means \(AI+BI+CI\) is minimized whenever \(AI=BI=CI\).

But that makes \(\triangle ABC\) equilateral!

I think you can take it from here. If you do the calculation, you'll see that the smallest value of \(AI+BI+CI\) is \(12\).

[I'm re-resharing this! Turns out I can do that. I'm doing this because I think these problems are cool and people should work on them. Specially on \(3\); I need a solution for that one. I suck at Number Theory :( ] Mursalin Habib · 3 years, 8 months ago

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The solution for \(3\) can be found here-

Solution to problem 3

PS- These are BdMO divisional problems and you will get faster responses from the BdMO forum for these problems. Labib Rashid · 3 years, 8 months ago

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@Labib Rashid Umm..that's not problem 3. Mursalin Habib · 3 years, 8 months ago

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@Mursalin Habib Messed the link up a bit. It's fixed now. Labib Rashid · 3 years, 8 months ago

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@Labib Rashid Thanks! Mursalin Habib · 3 years, 8 months ago

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