Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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Since <math>O_3R \parallel O_1O_2</math> and <math>\angle RO_3O_2 = \angle O_3O_2O_1</math>, then <math>O_2B \parallel O_1D</math>, and <math>BC \parallel AD</math> | Since <math>O_3R \parallel O_1O_2</math> and <math>\angle RO_3O_2 = \angle O_3O_2O_1</math>, then <math>O_2B \parallel O_1D</math>, and <math>BC \parallel AD</math> | ||
− | This means that <math>\Delta PDA \sim \Delta PBC \sim \Delta | + | This means that <math>\Delta PDA \sim \Delta PBC \sim \Delta O_3O_1O_1</math>. In other words, those three triangles are similar. |
Since <math>r_1</math> is the circumcenter of <math>\Delta PDA</math>, | Since <math>r_1</math> is the circumcenter of <math>\Delta PDA</math>, | ||
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Then, <math>A_1=\frac{|AD| \times h_1}{2}</math>, thus <math>h_1=\frac{2}{3}A_1</math> | Then, <math>A_1=\frac{|AD| \times h_1}{2}</math>, thus <math>h_1=\frac{2}{3}A_1</math> | ||
− | + | Since <math>PR</math> is the height of <math>\Delta O_1O_2O_3</math> to side <math>O_1O_2</math>, then using similar triangles, | |
− | Therefore, <math>\frac{r_1+r_2}{16}=\frac{3}{\frac{2}{3}A_1}</math> | + | <math>\frac{|O_1O_2|}{|PR|}=\frac{|AD|}{h_1}</math>. Therefore, <math>\frac{r_1+r_2}{16}=\frac{3}{\frac{2}{3}A_1}</math>. Solving for <math>r_2</math> we have: |
<math>r_2=\frac{72}{A_1}-r_1=\frac{72}{A_1}-\frac{18}{A_1}=\frac{54}{A_1}=3\left( \frac{18}{A_1} \right)=3r_1</math> | <math>r_2=\frac{72}{A_1}-r_1=\frac{72}{A_1}-\frac{18}{A_1}=\frac{54}{A_1}=3\left( \frac{18}{A_1} \right)=3r_1</math> |
Latest revision as of 01:05, 25 November 2023
Problem
Circles and are centered on opposite sides of line , and are both tangent to at . passes through , intersecting again at . Let and be the intersections of and , and and respectively. and are extended past and intersect and at and respectively. If and , then the area of triangle can be expressed as , where and are positive integers such that and are coprime and is not divisible by the square of any prime. Determine .
Solution
Let and be the centers of and respectively.
Let point be the midpoint of . Thus, and
Let and be the radii of circles and respectively.
Let and be the areas of triangles and respectively.
Since and , then , and
This means that . In other words, those three triangles are similar.
Since is the circumcenter of ,
then
Let be the height of to side
Then, , thus
Since is the height of to side , then using similar triangles,
. Therefore, . Solving for we have:
By similar triangles,
Using Heron's formula,
, where we have:
, thus
~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.
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |