Triangle ABC has area [ABC]=468. D,E and F are the midpoints of BC,CA and AB, respectively. Points P,Q and R are defined such that P is the incenter of AEF, Q is the incenter of BFD, and R is the incenter of CDE. What is [DREPFQ]?
Details and assumptions
[PQRS] denotes the area of figure PQRS.
The incenter of a triangle is the center of the circle which is inscribed in the triangle such that it is tangent to all 3 sides
Stop posting all these math problems without giving any thoughts of your own.
234
thanks
To find the area of the polygon DREPFQ, we need to consider the relationship between the areas of similar figures and their corresponding side lengths.
Since D, E, and F are the midpoints of the sides of triangle ABC, we can say that triangle DEF is similar to triangle ABC. This means that the ratio of the areas of triangle DEF to triangle ABC is equal to the square of the ratio of the corresponding side lengths.
Let's denote the side lengths of triangle DEF as x, y, and z, where x is DF, y is DE, and z is EF. The side lengths of triangle ABC are twice the lengths of triangle DEF, so we have the following relationships: AB = 2x, BC = 2y, and AC = 2z.
The area of triangle ABC is given as 468, so we have [ABC] = 468.
Now, we need to find the ratio of the lengths of the corresponding sides of the triangles. In triangle ABC, we can use the formula for the area of a triangle: [ABC] = (1/2) * BC * AP, where AP is the length of the perpendicular drawn from the incenter P to side BC.
Similarly, in triangle DEF, we have [DEF] = (1/2) * EF * DP.
Since triangle DEF is similar to triangle ABC, we can write:
[DEF] / [ABC] = (EF * DP) / (BC * AP)
Substituting the values we know, we get:
(1/468) * EF * DP = (1/2) * y * x
Simplifying, we have:
EF * DP = (234 * y * x)
Similarly, we can find the relationships for the other sides and lengths:
DF * EQ = (234 * x * z)
DE * FR = (234 * y * z)
Now we have the ratios of the lengths of the corresponding sides:
DP : BC :: EQ : AC :: FR : AB
Next, we can use the fact that the incenter is the center of the circle inscribed in the triangle, which is tangent to all three sides. This means that the angle bisectors of the triangle intersect at the incenter.
Since P is the incenter of triangle AEF, we can extend DP to intersect side EF at point S. Similarly, we can extend EQ and FR to intersect sides FD and DE at points T and U, respectively.
The polygon DREPFQ is formed by joining the points D, R, E, P, F, and Q in order.
Now, we can calculate the area of the polygon DREPFQ by calculating the areas of the triangles DRS, DEQ, and FQP and adding them together:
[DREPFQ] = [DRS] + [DEQ] + [FQP]
Using the formula for the area of a triangle, we have:
[DRS] = (1/2) * DS * SR
[DEQ] = (1/2) * EQ * QD
[FQP] = (1/2) * FQ * QP
Now, we have the ratios of the lengths of the sides:
DP : BC :: EQ : AC :: FR : AB
We can substitute these ratios into our calculated areas to find the final expression for [DREPFQ].