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
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 $2$:

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.

- 7 years, 6 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 :( ]

- 7 years, 6 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.

- 7 years, 6 months ago

Umm..that's not problem 3.

- 7 years, 6 months ago

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

- 7 years, 6 months ago

Thanks!

- 7 years, 6 months ago