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

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Try this beautiful problem from Geometry based on Centroid.

\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G_1\), \(G_2\), \(G_3\) respectively. The value of \(\frac{[G1G2G3]}{[ABC]}\) can BE represented by \(\frac{p}{q}\) for positive integers \(p\) and \(q\).

Find \(p+q\) where\([ABCD]\) denotes the area of ABCD.

- $14$
- $ 10$
- $7$

Geometry

Triangle

centroid

But try the problem first...

Answer: \(10\)

Source

Suggested Reading

Question Papers

Pre College Mathematics

First hint

\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G1\),\(G2\),\(G3\) respectively.we have to find out value of \(\frac{[G1G2G3]}{[ABC]}\) i.e area of \(\frac{[G1G2G3]}{[ABC]}\)

Let D, E, F be the midpoints of BC, CA, AB respectively.

Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$. Now can you find out \(GG_1,GG_2,GG_3\) ?

Can you now finish the problem ..........

Second Hint

we know that any median is divided at the centroid $2:1$

Now $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$

Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$

Thus, From homothetic transformation $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$

Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$

can you finish the problem........

Final Step

Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=$9+1$=\(10\)

- https://www.cheenta.com/problem-based-on-lcm-amc-8-2016-problem-20/
- https://www.youtube.com/watch?v=7AlfBAPWEMg

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?

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.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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