# USA 1996 Problem 5

Problem 5 of USA Mathematical Olympiad 1996

Triangle ABC has the following property: there is an interior point P such that ∠PAB = 10°, ∠PBA = 20°, ∠PCA = 30°, and ∠PAC = 40°. Prove that triangle ABC is isosceles.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Melvin Drimmer, Ph.D.) Extend CP to meet AB at S. From A draw a line to meet the extension of BP at R and CP at Q such that ∠QAP = 10°. We have BR = AR and

∠BRQ = 40°, ∠QAC = 30°, ∠AQC = 120°, ∠PQR = 60°, ∠QPR = ∠BSP = 80°, and ∠PSA = ∠QPB = 100°

We wish to prove that the triangles QSA and QPB are similar since, if they are, we have ∠QBP = ∠QAS = 20° and ∠QBA = 40°, ∠BQA = 120° = ∠BQC which cause the two triangles BQC and BQA to be congruent, implying BC = BA making triangle ABC isosceles.

In triangles QSA and QPB, ∠QSA = ∠QPB = 100°. So, to prove them similar, it suffices to show that QP / QS = PB / SA (1)

Since AP is bisector of ∠QAS, we have QP/QA = PS/SA (2)

QP/QA = PS/SA = QS/(QA + SA) or

QP/QS = QA/(QA + SA)

Combine with (1), we now need to prove

QA/(QA + SA) = PB/SA = (QA  PB)/QA = (QR + PR)/QA (since BR = AR),

or we need to prove

```			(QR + PR)/QA = PB/SA	(3)
```

Using the law of the sines, we have

QP/sin40° = QR/sin80° = PR/sin60° = (QR + PR)/(sin60° + sin80°) and PS/sin20° = PB/sin80°

or QP = (QR + PR)sin40°/(sin60° + sin80°) and PS = PB sin20°/sin80°

Substitute QP and PS to (2), it becomes

[(QR + PR)/QA ] x [sin40°/(sin60° + sin80°)] = [PB/SA] x [sin20°/sin80°]

so now we have to prove sin40°/(sin60° + sin80°) = sin20°/sin80° (4)

or sin20°/sin80° = (sin40° - sin20°)/sin60° or sin10°/sin30° = sin20°/sin80° or sin10°sin80° = sin30°sin20° or ½ (cos70° - cos90°) = cos60°cos70° or ½ = cos60° which is obvious!