Try this beautiful problem from the PRMO, 2018 based on Centroids and Area.

## Centroids and Area – PRMO 2018

Let ABC be an acute angled triangle and let H be its orthocentre. Let \(G_1\),\(G_2\) and \(G_3\) be the centroids of the triangles HBC, HCA, HAB. If area of triangle \(G_1G_2G_3\) =7 units, find area of triangle ABC.

- is 107
- is 63
- is 840
- cannot be determined from the given information

**Key Concepts**

Orthocentre

Centroids

Similarity

## Check the Answer

But try the problem first…

Answer: is 63.

PRMO, 2018, Question 21

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

AB=2DE in triangle \(HG_1G_2\) and triangle \(HDE\) \(\frac{AG_1}{HD}=\frac{G_1G_2}{DE}=\frac{2}{3}\) then \(G_1G_2=\frac{2DE}{3}=\frac{2AB}{3 \times 2}=\frac{AB}{3}\)

Second Hint

triangle \(G_1G_2G_3\) is similar triangle ABC then \(\frac{AreaatriangleABC}{Area G_1G_2G_3}=\frac{AB^{2}}{G_1G_2^{2}}=9\)

Final Step

then area triangle ABC=\(9 \times area triangle G_1G_2G_3\)=(9)(7)=63.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

## 2 replies on “Centroids and Area | PRMO 2018 | Question 21”

The orthocenter has no role in this problem it can be any intersection of cevians.I think so.

Yeah you are correct.

Google