+16 Let the incircle of △ABC be tangent to AB, BC, AC at points M, N, P, respectively. If ∡BAC = 30◦ , find [BP C] [ABC] · [BMC] [ABC] , where [ABC] denotes the area of △ABC.
+622 Here's how to find [BP C] [ABC] · [BMC] [ABC]:
Relate Areas using Tangent Lengths:
Since the incircle is tangent to the sides of the triangle at points M, N, and P, we know that AM, BN, and CP are radii of the incircle. Let the radius of the incircle be r.
The area of a triangle can be expressed using its base and height:
Area of ∆ABM = (AB)(AM)/2
Area of ∆BCN = (BC)(BN)/2
Area of ∆ACP = (AC)(CP)/2
Relate Base and Height using Angle Properties:
Since angle BAC = 30°, triangle ABC is a 30-60-90 triangle. This means:
AB = 2 * AC (where AC is the base of ∆ABM)
BC = AC√3 (where BC is the base of ∆BCN)
Substitute and Simplify:
Now, substitute the expressions for the bases from step 2 into the area equations from step 1:
Area of ∆ABM = (2 * AC)(r)/2 = AC * r
Area of ∆BCN = (AC√3)(r)/2 = (AC√3 * r)/2
Area Ratio:
We are interested in the ratio of the areas of triangles BPC and BMC:
[BPC] / [BMC] = (Area of ∆BCN) / (Area of ∆ABM) = [(AC√3 * r)/2] / (AC * r) = √3 / 2
Double Counting and Final Answer:
Notice that the area of triangle ABC appears twice in the product we need to find: once in the numerator ([BP C] [ABC]) and once in the denominator ([BMC] [ABC]).
This is because both triangles BPC and BMC share a base (BC) and height (altitude from A to BC) with triangle ABC.
Since the area of triangle ABC cancels out, we are left with:
[BP C] [ABC] · [BMC] [ABC] = √3 / 2 * √3 / 2 = (3/4)
Therefore, the product [BP C] [ABC] · [BMC] [ABC] is equal to 3/4.
