A similar question has been asked, but: A point P is chosen at random within right triangle ABC. Find the probability that the area of triangle PBC is less than a QUARTER of the area of triangle ABC.

TY

HelpPls123abc Jan 5, 2021

edited by
Guest
Jan 5, 2021

#1**+3 **

Since we can use any convenient right triangle...let's use a 3 - 4 - 5 right triangle where the legs are AB, AC and the hypotenuse is BC

The area = (1/2) (3)(4) = 6

(1/4)area = 6/4 = 3/2

Also....this area = (1/2) BC ( height)

3/2 = (1/2) (5) (height)

(3/2) /( 5/2) = 6/10 = 6/10 = 3/5 = height of PBC

See the following image :

Draw a perpendicular from A to BC gives us the altitude = 12/5 of the original triangle from A to BC

Let this altitude = AD

But.....we need an altitude of only 3/5

So at D = (1.44,1.92).....construct a circle of radius (3/5)

E = (1.08,1.44) = the point of tangency to the circle on a line parallel to BC

The equation of this line is y = (-3/4) ( x - 1.08)+ 1.44

And any P chosen on this line will form a triangle PBC that will have an area of 1/4 of ABC because the distance from this line to BC = 3/5

So any P between this line and BC will form a triangle that has an area < (1/4) [ABC ] because the altitude of this triangle PBC drawn to BC will be < 3/5

This line will intercept BA at F = (0, 2.25) and G at (3,0)

So....the area of GAF = (1/2) ( 3)(2.25) ≈ 3.375

So.....the probability that a "P" will be chosen such that PBC is < (1/4) [ABC] is

1 - area of GAF / area ABC =

1 - 3.375 / 6 = .4375 = 7/16

EDIT TO CORRECT A PREVIOUS MISTAKE

CPhill Jan 5, 2021