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Những bài giải cho kỳ thi toán Hoa Kỳ 2010

[ Thursday, 29 April, 2010 ]

Kỳ thi giỏi toán Hoa Kỳ năm nay 2010 đã xảy ra tuần qua. Đây là những bài gỉải.

Problem 1 of the United States Mathematical Olympiad 2010

Let AXYZB be a convex pentagon inscribed in a semicircle of diameter AB. Denote by P, Q, R, S the feet of the perpendiculars from Y onto lines AX, BX, AZ, BZ, respectively. Prove that the acute angle formed by lines PQ and RS is half the size of ∠XOZ, where O is the midpoint of segment AB.

Solution

Let AZ intercept BX at C, PQ and RS intercept at I. The acute angle formed by lines PQ and RS is

∠PIS = ∠PQY + ∠SRY – ∠QYR = ∠PQY + ∠SRY – (180° – ∠QCR) =
∠PQY + ∠SRY – ∠RCB

But ∠RCB subtends arcs AX and BZ; ∠PQY = ∠PXY subtends arc AY; ∠SRY = ∠SZY subtends arc BY.
Therefore, ∠PIS subtends arc AY + BY – AX – BZ = arc XZ = ½ ∠XOZ.

Problem 3 of the United States Mathematical Olympiad 2010

Let ABC be a triangle with ∠A = 90°. Points D and E lie on sides AC and AB, respectively, such that ∠ABD = ∠DBC and ∠ACE = ∠ECB. Segments BD and CE meet at I. Determine whether or not it is possible for segments AB, AC, BI, ID, CI, IE to all have integer lengths.

Solution

Apply the law of the cosine function, we have

BC² = BI² + CI² – 2 BI x CI cos∠BIC
But ∠BIC = 180° – ½ (180° – ∠A) = 135° and cos∠BIC = – ½ sqrt(2)
The above equation becomes BC² = BI² + CI² + sqrt(2) BI x CI
Or sqrt(2) BI x CI = BC² – BI² – CI²
Now assume that it is possible for segments AB, AC, BI, ID, CI and IE to all have integer lengths. BC² is then also an integer because BC² = AB² + AC² which, in turn, requires sqrt(2) BI x CI to be an integer. But since sqrt(2) is an irrational number, the product of sqrt(2) with an integer is not an integer. Therefore, our assumption was not possible, and it’s not possible for segments AB, AC, BI, ID, CI and IE to all have integer lengths.

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