What is the NO-SHORTCUT approach for learning great Mathematics?

# How to Pursue Mathematics after High School?

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

Try this beautiful Problem on Geometry based on Circle and Equilateral Triangle from AMC 10 A, 2017. You may use sequential hints to solve the problem.

## Circle and Equilateral Triangle  - AMC-10A, 2017- Problem 22

Sides $\overline{A B}$ and $\overline{A C}$ of equilateral triangle $A B C$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle A B C$ lies outside the circle? $?$

,

• $\frac{4 \sqrt{3} \pi}{27}-\frac{1}{3}$
• $\frac{\sqrt{3}}{2}-\frac{\pi}{8}$
• $12$
• $\sqrt{3}-\frac{2 \sqrt{3} \pi}{9}$
• $\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Geometry

Triangle

Circle

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2017 Problem-22

#### Check the answer here, but try the problem first

$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

## Try with Hints

#### First Hint

Given that ABC is a equilateral triangle whose $AB$ & $AC$ are the tangents of the circle whose centre is $O$. We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle

we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.

Later we have to find out red area and subtract from the Triangle ABC.

Now can you finish the problem?

#### Second Hint

Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$

Now Can you finish the Problem?

#### Third Hint

Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$

Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$

## What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

• What are some of the best colleges for Mathematics that you can aim to apply for after high school?
• How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
• What are the best universities for MS, MMath, and Ph.D. Programs in India?
• What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
• How can you pursue a Ph.D. in Mathematics outside India?
• What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

## Want to Explore Advanced Mathematics at Cheenta?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

To Explore and Experience Advanced Mathematics at Cheenta

This site uses Akismet to reduce spam. Learn how your comment data is processed.