Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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==Solution== | ==Solution== | ||
+ | |||
+ | [[File:AIME_2005_14a.png|600px]] | ||
Let <math>O_1, O_2,</math> and <math>O_3</math> be the centers of <math>\omega_1, \omega_2</math> and <math>\omega_3</math> respectively. | Let <math>O_1, O_2,</math> and <math>O_3</math> be the centers of <math>\omega_1, \omega_2</math> and <math>\omega_3</math> respectively. | ||
− | Let point <math>R</math> be the midpoint of <math>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{| | + | Let point <math>R</math> be the midpoint of <math>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math> |
Let <math>r_1</math> and <math>r_2</math> be the radii of circles <math>\omega_1</math> and <math>\omega_2</math> respectively. | Let <math>r_1</math> and <math>r_2</math> be the radii of circles <math>\omega_1</math> and <math>\omega_2</math> respectively. | ||
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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> | ||
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<math>A_1=\sqrt{\left( \frac{13}{2} \right)\left( \frac{13}{2}-3\right)\left( \frac{13}{2}-6 \right)\left( \frac{13}{2}-4 \right)}</math> | <math>A_1=\sqrt{\left( \frac{13}{2} \right)\left( \frac{13}{2}-3\right)\left( \frac{13}{2}-6 \right)\left( \frac{13}{2}-4 \right)}</math> | ||
+ | <math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</math> | ||
+ | <math>A_2=9A_1=\frac{9\sqrt{455}}{4}=\frac{p\sqrt{q}}{r}</math>, thus <math>p=9,q=455,r=4</math> | ||
+ | <math>p+q+r=\boxed{468}</math> | ||
− | + | ~Tomas Diaz. orders@tomasdiaz.com | |
− | + | {{alternate solutions}} | |
==See Also== | ==See Also== | ||
{{Mock AIME box|year=Pre 2005|n=3|num-b=13|num-a=15}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=13|num-a=15}} |
Latest revision as of 02:05, 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 |