Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 9"
Line 32: Line 32:
 
<math>A_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}</math>
 
<math>A_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}</math>
  
 +
<math>s=\frac{13+14+15}{2}=21</math>
 +
 +
<math>A_{ABC}=\sqrt{(21)(6)(7)(8)}=\sqrt{3^24^27^2}=(3)(7)(4)=\boxed{84}</math>
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 12:33, 26 November 2023

Problem

$ABC$ is a triangle with integer side lengths. Extend $\overline{AC}$ beyond $C$ to point $D$ such that $CD=120$. Similarly, extend $\overline{CB}$ beyond $B$ to point $E$ such that $BE=112$ and $\overline{BA}$ beyond $A$ to point $F$ such that $AF=104$. If triangles $CBD$, $BAE$, and $ACF$ all have the same area, what is the minimum possible area of triangle $ABC$?

Solution

Let "a", "b", and "c", be the lengths of sides $AB$, $BC$ and $CA$ respectively.

Let "h_a", "h_b", and "h_c", be the heights of $\Delta ABC$ from sides $AB$, $BC$ and $CA$ respectively.

Since the areas of triangles $CBD$, $BAE$, and $ACF$ are equal, then,

$104h_a=112h_b=120h_c$

Therefore,

$\frac{h_a}{h_b}=\frac{112}{104}=\frac{14}{13}$ and $\frac{h_a}{h_c}=\frac{120}{104}=\frac{15}{13}$

Since the area of $\Delta ABC$ is half any base times it's height, then:

$ah_a=bh_b=ch_c$

Therefore,

$b=\frac{h_a}{h_b}a=\frac{14}{13}a$ and $c=\frac{h_a}{h_c}a=\frac{15}{13}a$

Since "a", "b", and "c", and $13$ is a prime number, then the minimum integer value that $a$ can have in order for $b$ and $c$ to also be integer is $13$

Therefore $a=13$, $b=14$, and $c=15$

minimum possible area of triangle $ABC$ using Heron's formula is $ABC$ is:

$A_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$

$s=\frac{13+14+15}{2}=21$

$A_{ABC}=\sqrt{(21)(6)(7)(8)}=\sqrt{3^24^27^2}=(3)(7)(4)=\boxed{84}$

~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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_6_2006-2007_Problems/Problem_9&oldid=205888"