# APMO 2004 Problem 2

Problem 2 of Asian Pacific Mathematical Olympiad 2004

Let O be the circumcentre and H the orthocentre of an acute triangle ABC. Prove that the area of one of the triangles AOH, BOH and COH is equal to the sum of the areas of the other two.

Solutions by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Jim Krause)

Solution One From the vertices of triangle ABC A, B and C draw the altitudes to OH and intercept the extension of OH at M, G and P, respectively.

Since the three triangles AOH, BOH and COH have the same base OH, to prove the area of AOH is the sum of the areas of BOH and COH, we need to prove

 (1) GB + PC = AM

Extend CO to intercept the circle at D. Since CD is the diameter of the circle, we have ∠DAC = ∠DBC = ∠BFC = ∠AIC = 90°.

Or AD || HB and DB || AH; therefore, AD = HB

O and E are also midpoints of DC and AC, respectively, we have OE = AD/2

Or OE = HB/2

Let J be the midpoint of BH; from J draw the altitude to OH and cuts the extension of OH at K. We have

 (2) KJ = GB/2 HJ = HB/2 = OE and ∠KHJ = ∠OHF = ∠NOE

From E draw the altitude to OH and intercept it at N. The two triangles JKH and ENO are then congruent; we then have

 KJ = EN

Draw the line parallel to OH through E and intercepts AM and PC at L and Q, respectively; we then have

 KJ = EN = LM = QP (3) AL = QC

Combine with (2), GB + PC = 2 KJ + QC  QP

From (3), GB + PC = LM + QP + AL  QP

Or GB + PC = AM

which is the condition (1) we set out to prove.

Solution Two From the three vertices A, B and C of ΔABC draw lines ⊥ to OH and intercept it at M, G và P, respectively.

The three triangles AOH, BOH and COH share the same base OH, so to prove the areas of AOH to equal the areas of the other two we need to prove

```		AM = GB + PC
```

Let J be the midpoint of BC. AJ intercepts OH at N. From J draw the line to ⊥ and intercept OH at K. We see that GB + PC = 2JK. We then need to prove

```		AM = 2JK
```

Note that in a triangle, the three points centroid, orthocenter and circumcenter collinear; therefore, N is also the centroid of ΔABC

Therefore AN = 2NJ

or AM = 2JK because the two triangles AMN and JKN are similar.