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| | is the geometric mean of AD and DB if and only if | | is the geometric mean of AD and DB if and only if |
| | <math>\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})</math>. | | <math>\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})</math>. |
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| − | ==Solution==
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| − | Since this is an "if and only if" statement, we will prove it in two parts.
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| − | Before we begin, note a few basic but important facts.
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| − | 1. When two variables <math>x</math> and <math>y</math> are restricted by an equation <math>x+y=k</math> for some constant <math>k</math>, the maximum of their product occurs when <math>x=y=\frac{k}{2}</math>.
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| − | 2. The triangle inequality states that for a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> fulfills <math>a + b > c</math>, meaning that <math>a > \frac{c}{2}</math> or <math>b > \frac{c}{2}</math>, which is equivalent to saying that <math>a</math> and <math>b</math> both cannot be less than or equal to <math>\frac{c}{2}</math>.
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| − | 3. As point D moves along the base of the side AB, the locus of points where C can exist fills in a semicircle because C must be a distance <math>\sqrt{(AD)(DB)} = \sqrt{(\frac{AB}{2}-k)(\frac{AB}{2}+k)} = \sqrt{(\frac{AB}{2})^2-k^2}</math>
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| − | Part 1: If <math>\sin{A}\sin{B} \leq \sin^2{\frac{C}{2}}</math>, then such a point D exists.
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| − | First note that in order for D to exist, the line segment CD must be able to reach from the vertex C to the side AB, meaning that
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Revision as of 19:07, 6 January 2019
In the triangle ABC; prove that there is a point D on side AB such that CD
is the geometric mean of AD and DB if and only if
.
.
