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## Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Inscribed circle and perimeter.

## Inscribed circle and perimeter – AIME I, 1999

The inscribed circle of triangle ABC is tangent to AB at P, and its radius is 21 given that AP=23 and PB=27 find the perimeter of the triangle

• is 107
• is 345
• is 840
• cannot be determined from the given information

Inscribed circle

Perimeter

Triangle

## Check the Answer

AIME I, 1999, Question 12

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

Q tangency pt on AC, R tangency pt on BC AP=AQ=23 BP=BR=27 CQ=CR=x and

$s \times r =A$ and $s=\frac{27 \times 2+23 \times 2+x \times 2}{2}=50+x$ and A=$({(50+x)(x)(23)(27)})$ then from these equations 441(50+x)=621x then x=$\frac{245}{2}$

perimeter 2s=2(50+$\frac{245}{2}$)=345.

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## Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

### Key Concepts

Triangle Inequality

Perimeter

Geometry

## Check the Answer

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

## Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , $\frac {perimeter}{2} > a$

$\frac {perimeter}{2}$ = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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## What is Perimeter?

perimeter is a path that encompasses/surrounds a two-dimensional shape. It can be thought of as the length of the outline of a shape.

## Try the problem from AMC UP, 2014- Problem 11

These two squares, each with a side length of 10 cm, overlap as shown in the diagram. The shape of the overlap is also a square which has an area of 16 square centimeters. In centimeter s, what is the perimeter of the combined shape?

Australian Mathematics Competition (AMC UP), 2014, Problem No. 11

Perimeter

2 out of 10

Elementary Algebra by Hall and Knigh

## Use some hints

Find the Perimeter of 2 boxes

$(4 \times 10) \times 2$

$80 cm$

The perimeter of small box

$4 \times 4$

16 cm

So the remaining boxes Space perimeter is 80 – 16 = 64 cm

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## What is the area and perimeter of a circle?

A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

## Try the problem

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

AMC 8 2013 Problem 25

Geometry : Perimeter of a circle

7 out of 10

Mathematical Circles.

## Use some hints

First I want to give you the formula required.

You can clearly notice that we have to find the perimeters of all of the semicircles

The perimeter of a circle of radius $r$ unit can be obtained by the formula $2\pi r$. Then can you find perimeter of the semicircles ?!!!

So using the formula, the perimeters of

Semicircle 1 =$\frac{2\pi\times 100}{2}$ inches.

Semicircle 2 =$\frac{2\pi\times 60}{2}$ inches.

Semicircle 3 =$\frac{2\pi\times 80}{2}$ inches.

So the total path covered by the ball is

$\pi(100+60+80)=240\pi$ inches.

Is it the final answer??? Or have we ignored something ?

OK !!! please notice that they have asked for the distance covered by the center of the ball.

And the ball is of radius $2$ inches.

So for the $1^{st}$ and $3^{rd}$ semicircle : The center will roll along a semicircular path of radius $R_1-2$ and $R_3-2$.

See this image :

And for the $2^{nd}$ semicircle : The center will roll along a semicircular path of radius $R_2+2$.

See the image below :

So the length of the path covered by the center of the ball is

$[\pi(100-2)+\pi(60+2)+\pi(80-2)] \quad \text{inches} \\=\pi(98+62+78) \quad \text{inches}\\=238\pi \quad \text{inches}$.