Difference between revisions of "2006 iTest Problems/Problem 32"
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==Solution 2 (Faster and less Calculation)== | ==Solution 2 (Faster and less Calculation)== | ||
− | Draw line <math>\overline{AD}</math> where it is a bisector of <math>\angle PAQ</math> Then you can use angle bisctors <math>\overline{AP}</math>, <math>\overline{AD}</math>, and <math>\overline{AQ}</math> to get the following ratios: | + | Draw line <math>\overline{AD}</math> where it is a bisector of <math>\angle PAQ</math>. Let <math>\overline{PD}</math> be x. Then you can use angle bisctors <math>\overline{AP}</math>, <math>\overline{AD}</math>, and <math>\overline{AQ}</math> to get the following ratios: |
− | + | <cmath>\frac{\overline{AB}}{\overline{AQ}} = \frac{3}{5}</cmath> | |
− | + | <cmath>\frac{\overline{AP}}{\overline{AC}} = \frac{7}{20}</cmath> | |
+ | <cmath>\frac{\overline{AQ}}{\overline{AP}} = \frac{35-x}{x}</cmath> | ||
+ | |||
+ | Multiplying the ratios, we can find that <math>\frac{\overline{AB}}{\overline{AC}} = \frac{21(35-x)}{100x}</math>. On the other hand, using <math>\overline{AD}</math> once again on the big triangle <math>\triangle{ABC}</math>, we find that <math>\frac{\overline{AB}}{\overline{AC}}</math> also equals to <math>\frac{21+x}{135-x}</math>. If we try to calculate this directly, the numbers would be astronomically high. Therefore, we can introduce k. Let k be <math>\frac{\overline{AB}}{\overline{AC}}</math>. Then, the substitution for x in terms of k: <math>\frac{135k-21}{k+1}</math>. Plugging k in, we get: | ||
+ | <cmath>k = \frac {21(35-\frac{135k-21}{k+1})} {\frac{100(135k-21)}{k+1}}</cmath> | ||
+ | |||
+ | Simplyfing, we cancel out the <math>2100</math>k's, and we get <cmath>13500k^2 = 21\cdot56</cmath> | ||
+ | Solving for k, we get <math>k = \frac{7\sqrt{10}}{75}</math> | ||
+ | |||
+ | Then adding <math>7</math>, <math>10</math>, and <math>75</math>, we get <math>\boxed {92}</math>, as desired. | ||
+ | |||
+ | ~CC2010CC2015 | ||
+ | |||
+ | NOTE: NO TRIG REQUIRED BTW ANOTHER WAY IS LAW OF SINES | ||
==See Also== | ==See Also== |
Latest revision as of 21:49, 29 July 2025
Problem
Triangle is scalene. Points
and
are on segment
with
between
and
such that
,
, and
. If
and
trisect
, then
can be written uniquely as
, where
and
are relatively prime positive integers and
is a positive integer not divisible by the square of any prime. Determine
.
Solution
Let and
. Since
, by the Angle Bisector Theorem, we have
and
.
By using the Law of Cosines on and
, we have
By using the Law of Cosines on
and
, we have
Multiplying the second equation by
and adding the two equations results in
After substituting
back, solve for
to get
Thus,
, so
.
NOTE: SIMPLY USE STEWARTS THEOREM
Solution 2 (Faster and less Calculation)
Draw line where it is a bisector of
. Let
be x. Then you can use angle bisctors
,
, and
to get the following ratios:
Multiplying the ratios, we can find that . On the other hand, using
once again on the big triangle
, we find that
also equals to
. If we try to calculate this directly, the numbers would be astronomically high. Therefore, we can introduce k. Let k be
. Then, the substitution for x in terms of k:
. Plugging k in, we get:
Simplyfing, we cancel out the k's, and we get
Solving for k, we get
Then adding ,
, and
, we get
, as desired.
~CC2010CC2015
NOTE: NO TRIG REQUIRED BTW ANOTHER WAY IS LAW OF SINES
See Also
2006 iTest (Problems, Answer Key) | ||
Preceded by: Problem 31 |
Followed by: Problem 33 | |
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