Difference between revisions of "2005 Canadian MO Problems/Problem 4"
(A better incomplete solution.) |
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− | == | + | ==Incomplete Solution== |
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Let the sides of triangle <math>ABC</math> be <math>a</math>, <math>b</math>, and <math>c</math>. Thus <math>\dfrac{abc}{4K}=R</math>, and <math>a+b+c=P</math>. We plug these in: | Let the sides of triangle <math>ABC</math> be <math>a</math>, <math>b</math>, and <math>c</math>. Thus <math>\dfrac{abc}{4K}=R</math>, and <math>a+b+c=P</math>. We plug these in: | ||
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<cmath>\dfrac{KP}{R^3}=\dfrac{(a+b+c)^3(-a+b+c)^2(a-b+c)^2(a+b-c)^2}{4a^3b^3c^3}</cmath> | <cmath>\dfrac{KP}{R^3}=\dfrac{(a+b+c)^3(-a+b+c)^2(a-b+c)^2(a+b-c)^2}{4a^3b^3c^3}</cmath> | ||
− | + | ==Solution Outline== | |
+ | ^hahahaha... you can probably use Ravi Sub. to finish the above. | ||
− | + | OR | |
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− | + | Use the formula <math>K=\dfrac{abc}{4R}</math> to get <math>KP/R^3=\dfrac{abc(a+b+c)}{4R^4}</math>. Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish. (Jensen's is used for <math>\sin A+\sin B+\sin C \le \dfrac{3\sqrt3}{2}</math>. |
Revision as of 13:37, 25 October 2013
Incomplete Solution
Let the sides of triangle be
,
, and
. Thus
, and
. We plug these in:
.
Now Heron's formula states that . Thus,
Solution Outline
^hahahaha... you can probably use Ravi Sub. to finish the above.
OR
Use the formula to get
. Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish. (Jensen's is used for
.