Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 9"
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<math>b=\frac{h_a}{h_b}a=\frac{14}{13}a</math> and <math>c=\frac{h_a}{h_c}a=\frac{15}{13}a</math> | <math>b=\frac{h_a}{h_b}a=\frac{14}{13}a</math> and <math>c=\frac{h_a}{h_c}a=\frac{15}{13}a</math> | ||
+ | |||
+ | Since "a", "b", and "c", and <math>13</math> is a prime number, then the minimum integer value that <math>a</math> can have in order for <math>b</math> and <math>c</math> to also be integer is <math>13</math> | ||
+ | |||
+ | Therefore <math>a=13</math>, <math>b=14</math>, and <math>c=15</math> | ||
+ | |||
+ | minimum possible area of triangle <math>ABC</math> using Heron's formula is <math>ABC</math> is: | ||
+ | |||
+ | <math>A_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}</math> | ||
+ | |||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 13:31, 26 November 2023
Problem
is a triangle with integer side lengths. Extend
beyond
to point
such that
. Similarly, extend
beyond
to point
such that
and
beyond
to point
such that
. If triangles
,
, and
all have the same area, what is the minimum possible area of triangle
?
Solution
Let "a", "b", and "c", be the lengths of sides ,
and
respectively.
Let "h_a", "h_b", and "h_c", be the heights of from sides
,
and
respectively.
Since the areas of triangles ,
, and
are equal, then,
Therefore,
and
Since the area of is half any base times it's height, then:
Therefore,
and
Since "a", "b", and "c", and is a prime number, then the minimum integer value that
can have in order for
and
to also be integer is
Therefore ,
, and
minimum possible area of triangle using Heron's formula is
is:
, where
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.