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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 88 and area 1212. What is the radius of the largest circle that can be inscribed in this triangle?
  3. The least consecutive ten numbers, all greater than 1010, are determined that are respectively divisible by the numbers 1 through 10. Write down the smallest number among these 1010.
  4. In trapezium ABCD,ADBC,AD<BCABCD, AD||BC, AD < BC, unparallel sides are equal. A circle with centre O is inscribed in the trapezium. OADOAD is equilateral. Find the radius of the circle if the area of the trapezium is 8003\frac{800}{\sqrt{3}}

Note by Tahmid Kenshin
7 years, 6 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 22:

You can immediately see that the height of the triangle is 33. You should be familiar with the relation, rs=[ABC]rs=[ABC] where rr and ss are the inradius and semi perimeter of ABC\triangle ABC and [ABC][ABC] denotes its area.

So, in order to maximize rr, we need to minimize the perimeter.

Now draw two parallel lines l1l_1 and l2l_2 such that they are 33 units away from each other. Set points AA and BB on l1l_1 such that AB=8AB=8. Now notice that any point CC on l2l_2 satisfies the criteria of a triangle with base 88 and area 1212.

Our job is to minimize AC+BCAC+BC.

How do we do that?

If you know calculus, you can put in some co-ordinates for points AA, BB and CC and differentiate AC+BCAC+BC and set it to zero. I'm going to take a purely geometrical approach.

Let AA' be the reflection of AA over l2l_2. Now AC+BC=AC+BCAC+BC=A'C+BC and AC+BCA'C + BC is minimized whenever CC is on the line ABA'B [triangle inequality]. Now if CC is on ABA'B, CC is the midpoint of ABA'B. So, AC=BCA'C=BC and that means ACAC has to be equal to BCBC if you want to minimizeAC+BCAC+BC . I think you can take it from here. I'm skipping ahead of the calculation. The answer is 43\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 - 7 years, 6 months ago

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

Solution to 11:

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 joinA,IA, I; B,IB, I and C,IC, I, you're going to get three triangles with height r=2r=2 and the sides of ABC\triangle ABC as their bases.

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

So, AI+BIAI+BI is minimized if and only if AI=BIAI=BI.

BI+CIBI+CI is minimized if and only if BI=CIBI=CI.

And, CI+AICI+AI is minimized if and only if CI=AICI=AI.

That means AI+BI+CIAI+BI+CI is minimized whenever AI=BI=CIAI=BI=CI.

But that makes ABC\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+CIAI+BI+CI is 1212.

[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 33; I need a solution for that one. I suck at Number Theory :( ]

Mursalin Habib - 7 years, 6 months ago

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The solution for 33 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 - 7 years, 6 months ago

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Umm..that's not problem 3.

Mursalin Habib - 7 years, 6 months ago

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Messed the link up a bit. It's fixed now.

Labib Rashid - 7 years, 6 months ago

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@Labib Rashid Thanks!

Mursalin Habib - 7 years, 6 months ago

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