Points p , q , r are taken on BC CA AB such that Bp/ Pc = Cq/qA = Ar/rB = m/n. Find the ratio of area of Triangle ABC and pqr

Question

You should also give the ratio or the answer   which is contained in m and n only. Please give the answer quickly and pqr is a triangle 

Answer 1

There is probably some short simple brilliant solution to this one, but here's the approach I would take using a bit of trigonometry:

 

Note that the area of the large triangle = AB*AC*sin A = AB*BC*sin B = AC*BC*sin C.

 

Note that area of triangle Aqr = Ar*Aq*sin A = (n * AB / (m+n))  * (m*AC / (m+n)) * sin A = nm * AB * AC * sin A / (m+n)²

 

Computing the area of the triangles Brp and Cqp, the same way, note that each of these three has the same area.  Therefore, the area of the small triangle is:

 

AB*AC*sin A - 3 * nm * AB * AC * sin A / (m+n)²

 

So Area ABC / Area pqr = AB * AC * sin A / (AB*AC*sin A - 3 * nm * AB * AC * sin A / (m+n)²)

 

Note that AB * AC * sin A cancels out, so we get:

 

1 / (1 - 3 * nm /(m+n)²) = (m+n)² / [(m+n)² - 3 * mn]

 

Note that in the special case where m = n, this gives us 4, as expected.  In the special case where m = 0, this gives us 1, also as expected.

 

There's probably another tutor on here with a more elegant solution, but hopefully this helps you.  If you have additional questions, feel free to let me know.

Answer 2

In other words, P, Q & R are midpoints of the respective sides of triangle ABC.

 

Consequently area of triangle PQR will be (½)² of the area of triangle ABC, i.e ratio of small triangle's area to large triangle's area will be 1:4.