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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, 11 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.

- 3 years, 11 months ago

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 :( ]

- 3 years, 11 months ago

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.

- 3 years, 11 months ago

Umm..that's not problem 3.

- 3 years, 11 months ago

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

- 3 years, 10 months ago

Thanks!

- 3 years, 10 months ago