大师作文网一道IMO试题Construct a right-angled triangle whose hypotenuse c
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一道IMO试题
Construct a right-angled triangle whose hypotenuse c is given
if it is known that the median from the right angle equals the geometric
mean of the remaining two sides of the triangle
Construct a right-angled triangle whose hypotenuse c is given
if it is known that the median from the right angle equals the geometric
mean of the remaining two sides of the triangle
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Analysis.Let a and b be the other two sides of the triangle.From the
conditions of the problem we have c2 = a2 +b2 and c/2=
√
ab ⇔ 3/2c2 =
a2+b2+2ab =(a+b)2 ⇔ 3/2c = a+b.Given a desired ABC let D be
a point on (AC such that CD = CB.In that case,AD = a+b = 3/2c,
and also,since BC = CD,it follows that ∠ADB =45◦.
Construction.From a segment of length c we elementarily construct a
segment AD of length 3/2 c.Wethenconstructaray(DX such that
∠ADX =45◦ and a circle k(A,c) that intersects the ray at point B.
Finally,we construct the perpendicular from B to AD;point C is the
foot of that perpendicular.
Proof.It holds that AB = c,and,since CB = CD,it also holds that AC+
CB = AC +CD = AD = 3/2 c.From this it follows that
√
AC • CB =
c/2.Since BC is perpendicular to AD,it follows that BCA =90◦.Thus
ABC is the desired triangle.
Discussion.Since AB
√2=
√2c> 3/2 c = AD > AB,thecircle k
intersects the ray DX in exactly two points,which correspond to two
symmetric solutions.
Analysis.Let a and b be the other two sides of the triangle.From the
conditions of the problem we have c2 = a2 +b2 and c/2=
√
ab ⇔ 3/2c2 =
a2+b2+2ab =(a+b)2 ⇔ 3/2c = a+b.Given a desired ABC let D be
a point on (AC such that CD = CB.In that case,AD = a+b = 3/2c,
and also,since BC = CD,it follows that ∠ADB =45◦.
Construction.From a segment of length c we elementarily construct a
segment AD of length 3/2 c.Wethenconstructaray(DX such that
∠ADX =45◦ and a circle k(A,c) that intersects the ray at point B.
Finally,we construct the perpendicular from B to AD;point C is the
foot of that perpendicular.
Proof.It holds that AB = c,and,since CB = CD,it also holds that AC+
CB = AC +CD = AD = 3/2 c.From this it follows that
√
AC • CB =
c/2.Since BC is perpendicular to AD,it follows that BCA =90◦.Thus
ABC is the desired triangle.
Discussion.Since AB
√2=
√2c> 3/2 c = AD > AB,thecircle k
intersects the ray DX in exactly two points,which correspond to two
symmetric solutions.
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