Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 5"
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==Solution== | ==Solution== | ||
+ | |||
+ | ===Solution 1=== | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.15cm); | ||
+ | draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle); | ||
+ | draw((0,24sqrt(2))--(8,-8sqrt(2))); | ||
+ | draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0))); | ||
+ | draw((8,-8sqrt(2))--(-12,0)); | ||
+ | draw((8,-8sqrt(2))--(28,0)); | ||
+ | draw((8,-8sqrt(2))--(8,0)); | ||
+ | label("$A$",(0,24sqrt(2)),NNW); | ||
+ | label("$B$",(-12,0),WSW); | ||
+ | label("$C$",(28,0),ESE); | ||
+ | label("$D$",(6,0),NW); | ||
+ | label("$H$",(8,0),NNE); | ||
+ | label("$E$",(8,-8sqrt(2)),S); | ||
+ | </asy> | ||
<math>\angle BAE \cong \angle BCE</math> because they are both subscribed by arc <math>BE</math>. <math>\angle CAE \cong \angle CBE</math> because they are both subscribed by arc <math>CE</math>. Hence <math>\angle BCE \cong \angle CBE</math>, because <math>\angle BAD \cong CAD</math>. Then <math>\Delta BEC</math> is isosceles. | <math>\angle BAE \cong \angle BCE</math> because they are both subscribed by arc <math>BE</math>. <math>\angle CAE \cong \angle CBE</math> because they are both subscribed by arc <math>CE</math>. Hence <math>\angle BCE \cong \angle CBE</math>, because <math>\angle BAD \cong CAD</math>. Then <math>\Delta BEC</math> is isosceles. | ||
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Then by the Pythagorean Theorem on <math>\Delta DHE</math>, <math>4+HE^2=DE^2</math>. Also from <math>\Delta CHE</math>, <math>400+HE^2=CE^2=4DE^2</math>. Subtracting these equations yields <math>396=3DE^2</math>, and so <math>DE^2=\boxed{132}</math>. | Then by the Pythagorean Theorem on <math>\Delta DHE</math>, <math>4+HE^2=DE^2</math>. Also from <math>\Delta CHE</math>, <math>400+HE^2=CE^2=4DE^2</math>. Subtracting these equations yields <math>396=3DE^2</math>, and so <math>DE^2=\boxed{132}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | <asy> | ||
+ | unitsize(0.15cm); | ||
+ | draw((28,0)--(0,24sqrt(2))--(-12,0)--cycle); | ||
+ | draw((0,24sqrt(2))--(8,-8sqrt(2))); | ||
+ | draw(circumcircle((28,0),(0,24sqrt(2)),(-12,0))); | ||
+ | label("$A$",(0,24sqrt(2)),NNW); | ||
+ | label("$B$",(-12,0),WSW); | ||
+ | label("$C$",(28,0),ESE); | ||
+ | label("$D$",(6,0),NW); | ||
+ | label("$E$",(8,-8sqrt(2)),S); | ||
+ | </asy> | ||
+ | |||
+ | Let <math> BD=x </math>, so that <math> DC=40-x </math>. From the Angle Bisector Theorem, <math> \frac{x}{36}=\frac{40-x}{44} </math>. Cross-multilplying and solving for <math> x </math>, we find that <math> x=18 </math>. Thus, <math> BD=18 </math> and <math> DC=22 </math>. | ||
+ | |||
+ | |||
+ | Now, from Stewart's Theorem, <math> AB^2\cdot CD+AC^2\cdot BD=AD^2\cdot BC+BC\cdot BD\cdot CD </math>. Plugging in values and letting <math> AD=y </math>, we find that <math> 36^2\cdot22+44^2\cdot18-18\cdot22\cdot40=40y^2 </math>. | ||
+ | |||
+ | |||
+ | Dividing both sides by <math> 40 </math> gives <math> \frac{18^2\cdot22+22^2\cdot18}{10}-18\cdot22=y^2 </math>.Factoring a <math> 18\cdot22 </math> out of the numerator of the fraction and continuing to simplify, we find that <math> y^2=18\cdot22\cdot3 </math>. | ||
+ | |||
+ | |||
+ | Now, from Power of a Point on <math> D </math>, we have <math> BD^2\cdot DC^2=AD^2\cdot DE^2 </math>. Now, let <math> DE=z </math>, so we have <math> (18\cdot22)^2=(18\cdot22\cdot3)z^2 </math>. From here, we can find that <math> z^2=\boxed{132} </math>. |
Revision as of 23:09, 1 January 2012
Contents
Problem
In triangle
The bisector of angle
meet
at
and the circumcircle at
different from
. Calculate the value of
Solution
Solution 1
because they are both subscribed by arc
.
because they are both subscribed by arc
. Hence
, because
. Then
is isosceles.
Let be the foot of the perpendicular from
to
. As
is isosceles, it follows that
is the midpoint of
, and so
. From the angle bisector theorem,
. We have
. Solving this system of equations yields
. Thus,
.
because they are vertical angles. It was shown
, and so
by
similarity. Then
and so
.
Then by the Pythagorean Theorem on ,
. Also from
,
. Subtracting these equations yields
, and so
.
Solution 2
Let , so that
. From the Angle Bisector Theorem,
. Cross-multilplying and solving for
, we find that
. Thus,
and
.
Now, from Stewart's Theorem, . Plugging in values and letting
, we find that
.
Dividing both sides by gives
.Factoring a
out of the numerator of the fraction and continuing to simplify, we find that
.
Now, from Power of a Point on , we have
. Now, let
, so we have
. From here, we can find that
.