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 12:31, 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.
