# APMO 1989 Problem 3

Problem 3 of Asian Pacific Mathematical Olympiad 1989

Let A1, A2, A3 be three points in the plane, and for convenience, let A4 = A1, A5 = A2. For n = 1, 2, and 3, suppose that Bn is the midpoint of AnAn+1, and suppose that Cn is the midpoint of AnBn. Suppose that AnCn+1 and BnAn+2 meet at Dn, and that AnBn+1 and CnAn+2 meet at En. Calculate the ratio of the area of triangle D1D2D3 to the area of triangle E1E2E3.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Connect and extend A1D3 to meet A2A3 at A, A2D1 to meet A1A3 at B and A3D2 to meet A1A2 at C.

Apply Ceva's theorem for the three lines A1C2, A3B1 and A2B, we have:

{A2C2 \over A3C2}×{A3B \over A1B}×{A1B1 \over A2B1} = 1

since A3C2 = 3×A2C2 and A1B1 = A2B1, we have

{A1B \over A3B} = {1 \over 3}

We also have

{A1C1 \over A2C1} = {1 \over 3 }

Therefore BC1 || A2A3 and

{C1B \over A2A3} = {A1C1 \over A1A2} = {1 \over 4}

We then have

{A2B2 \over A3B2}×{A3B \over A1B}×{A1C1 \over A2C1} = 1 × 3×{1 \over3} = 1

therefore, per Ceva's theorem point E1 is on A2B.

With the same argument, E2 is on A3C and E3 is on A1A.

Now apply Ceva's theorem for the three lines E1B2, A2D3 and A3D1 that meet at G, we have

{E1D3 \over A3D3} = {E1D1 \over A2D1}
or D1D3 || A2A3

Similarly, D2D3 || A1A2 and D1D2 || A1A3

and for the three lines GB2, A2E3 and A3E2 that meet at D2, we have

{GE3 \over A3E3} = {GE2 \over A2E2}
or E2E3 || A2A3

Similarly, E1E2 || A1A2 and E1E3 || A1A3

ΔD1D2D3 and ΔE1E2E3 are similar since their corresponding sides are parallel to each other.

With the parallel lines, we now have

{E1D1 \over E1A2} = {D1D3 \over A2A3} = {D1D3 \over 2AC2} = {A1D1 \over 2A1C2} = {A2D1 \over 2A2B}

or

{A2D1 \over 2E1D1} = {A2B \over E1A2} = {E1A2 + E1B \over E1A2} = 1+ {E1B \over E1A2} = 1 + {C1B \over A2A3} = 1 + {1 \over 4} = {5 \over 4}

or

{2E1D1 \over A2D1} = {2E1D3 \over A3D3} = {4 \over 5}
or
{E1D3 \over A3D3} = {2 \over 5}

Add 1 to both sides, we have

1 + {E1D3 \over A3D3} = {7 \over 5}
or
{A3D3 + E1D3 \over A3D3} = {7 \over 5}
or
{E1A3 \over A3D3} = {7 \over 5}
or
{A3D3 \over E1A3} = {5 \over 7}

or

{D2D3 \over E2E1} = {5 \over 7}

Thus the ratio of the corresponding sides of the two similar triangles ΔD1D2D3 and ΔE1E2E3 is equal to 5/7.

Therefore the ratio of the area of ΔD1D2D3 to the area of Δ E1E2E3 is

{5 \over 7}×{5 \over 7} = {25 \over 49}