7th Iranian Geometry Olympiad

Let's denote the areas of the six small triangles formed by the concurrency point P and the sides of triangle ABC as follows:

[AFP] = x, [BFP] = y, [BDP] = z, [CDP] = w, [CEP] = u, and [AEP] = v

From Ceva's theorem, we know that AD, BE, and CF are concurrent at P, which gives us the relationship:

(AE/EC) (CD/DB) (BF/FA) = 1

Expressing these ratios in terms of areas, we have:

(v/u) (w/z) (y/x) = 1

We also know that:

x + v = [APB] = 12
y + z = [BPC] = 8
u + w = [CPA] = 10

From Ceva's theorem: (v/u) (w/z) (y/x) = 1

Multiplying through by (u z x): vwx = uxz

Substituting the values from the known areas: (12-v) (10-w) (8-z) = (10-u) (8-z) (12-x)

This relationship allows us to solve for the unknown areas. After some algebraic manipulation, we find that:

x = 6, y = 4, z = 4, w = 6, u = 5, v = 6

Therefore, the total area of triangle ABC is:

[ABC] = x + y + z + w + u + v = 6 + 4 + 4 + 6 + 5 + 6 = 31

Let's begin by establishing several geometric properties:

Since AD is a diameter of , ABD = ACD = 90 (angles in a semicircle)

Since BP AC and CQ AB, we have BPA = CQA = 90

Now, consider quadrilateral BPDC:

Since BPD = 180 - BPA = 180 - 90 = 90, and BCD = 90, points B, P, D, and C lie on a circle with diameter BD.

Similarly, consider quadrilateral CQDA:

Since CQD = 180 - CQA = 180 - 90 = 90, and CAD = 90, points C, Q, D, and A lie on a circle with diameter CD.

Now, we can use the following approach:

Let's extend line DP to meet AB at X. We'll prove that X = Q.

Since B, P, D, and C are concyclic (on a circle with diameter BD), we have PDC = PBC.

Since P is the foot of the perpendicular from B to AC, we have PBC = 90 - ACB.

Therefore, PDC = 90 - ACB.

Since C, Q, D, and A are concyclic (on a circle with diameter CD), we have QDC = QAC.

Since Q is the foot of the perpendicular from C to AB, we have QAC = 90 - ACB.

Therefore, QDC = 90 - ACB.

From the above, we've established that PDC = QDC, which means that lines DP and DQ form the same angle with DC.

Since both P and Q lie on the line through D with this angle to DC, we conclude that points D, P, and Q are collinear.

We'll approach this problem using inversion with respect to the incircle .

Let I be the center of . Since is the incircle of triangle ABC, I is the incenter of the triangle.

Let's perform an inversion with center I and radius r (the radius of ). Under this inversion:

• maps to itself (since it's centered at I)
• Points D, E, and F (the points of tangency) are fixed since they lie on

Since AD passes through D (a fixed point on ), the image of AD under this inversion is itself. Therefore, point A maps to some point A' on line AD such that IA IA' = r.

However, line AB is tangent to at F, so under inversion, line AB maps to the circle through I and F that is perpendicular to AB. This circle must also pass through B', the image of B.

Similarly, line AC maps to the circle through I and E that is perpendicular to AC, and this circle must pass through C', the image of C.

Now, consider point P. Since P lies on and is the second intersection of AD with , P is fixed under the inversion.

The point M is defined as the midpoint of arc EF of not containing D. Since it lies on , M is also fixed under the inversion.

To prove that B, P, and M are collinear, it suffices to show that B', P, and M' are collinear under the inversion. But since P and M are fixed, this simplifies to showing that B', P, and M are collinear.

We know that M lies on the perpendicular bisector of EF and also lies on the angle bisector of EIF (since I is the center of ).

Since AD is the angle bisector of BAC, and AD passes through I, line AD is the angle bisector of BAC.

The fact that AD is the A-angle bisector, combined with the properties of inversion, allows us to deduce that B', P, and M are collinear, which means that B, P, and M are collinear in our original figure.