Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 9"
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==Solution== | ==Solution== | ||
+ | [[File:MockAIME_6_P9.png|600px]] | ||
− | Let | + | Let <math>a</math>, <math>b</math>, and <math>c</math>, be the lengths of sides <math>AB</math>, <math>BC</math> and <math>CA</math> respectively. |
− | Let | + | Let <math>h_a</math>, <math>h_b</math>, and <math>h_c</math>, be the heights of <math>\Delta ABC</math> from sides <math>AB</math>, <math>BC</math> and <math>CA</math> respectively. |
Since the areas of triangles <math>CBD</math>, <math>BAE</math>, and <math>ACF</math> are equal, then, | Since the areas of triangles <math>CBD</math>, <math>BAE</math>, and <math>ACF</math> are equal, then, | ||
Line 24: | Line 25: | ||
<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 | + | Since <math>a</math>, <math>b</math>, and <math>c</math>, are integers, 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> | Therefore <math>a=13</math>, <math>b=14</math>, and <math>c=15</math> |
Revision as of 13:46, 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 ,
, and
, be the lengths of sides
,
and
respectively.
Let ,
, and
, 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 ,
, and
, are integers, 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.