- The incenter of triangle \(ABC\) is \(I\) and inradius is \(2\). What is the smallest possible value of \(AI+BI+CI\) ?
- 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?
- 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 \(10\).
- 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 \(\frac{800}{\sqrt{3}}\)

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TopNewestI 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 ago

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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 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 ago

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– Mursalin Habib · 3 years ago

Umm..that's not problem 3.Log in to reply

– Labib Rashid · 3 years ago

Messed the link up a bit. It's fixed now.Log in to reply

– Mursalin Habib · 3 years ago

Thanks!Log in to reply