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Bài thi giỏi toán Hoa Kỳ năm 1996

[ Friday, 23 April, 2010 ]

Problem 5 of USA Mathematical Olympiad 1996

As always it’s in English for international students to search.

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 Vo Duc Dien

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°,
∠PSA = ∠QPB = 100°, and ∠BSP = ∠QPR = 80°

We need to prove the two triangles QSA and QPB are similar since if they are, we will have ∠QBP = ∠QAS = 20° and ∠QBA = 40°, ∠BQA = 120° = ∠BQC and the two triangles BQC and BQA are congruent and thus BC = BA and the triangle ABC is isosceles.

To prove those two triangles QSA and QPB that already have the ∠QSA = ∠QPB = 100° similar, we need to prove 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 since function, 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! Q.E.D

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