Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 5"
(Created page with "==Problem== In triangle <math>ABC,</math> <math>AB=36,BC=40,CA=44.</math> The bisector of angle <math>A</math> meet <math>BC</math> at <math>D</math> and the circumcircle at <ma...") |
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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. | ||
− | Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math> , and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>. Thus, <math>DH=CD-CH=22-20=2</math>. | + | Let <math>H</math> be the foot of the perpendicular from <math>E</math> to <math>BC</math>. As <math>\Delta BEC</math> is isosceles, it follows that <math>H</math> is the midpoint of <math>BC</math>, and so <math>HC=20</math>. From the angle bisector theorem, <math>\frac{36}{BD}=\frac{44}{CD}</math>. We have <math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>. Thus, <math>DH=CD-CH=22-20=2</math>. |
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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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | This is the author's solution. Refer to the above diagrams. | ||
+ | |||
+ | Notice that | ||
+ | <cmath>\angle AEC=\angle ABD</cmath> | ||
+ | <cmath>\angle EAC =\angle BAD</cmath> | ||
+ | So by AA similarity, <math>\triangle AEC \sim \triangle ABD.</math> It follows that <math>\frac{AE}{AC} = \frac{AB} {AD},</math> or <math>AE*AD= AB*AC.</math> | ||
+ | Now let <math>AD=x, DE=y.</math> By the previous result, we have | ||
+ | <cmath>x(x+y)=x^2+xy=36*44=16*99</cmath> | ||
+ | But by the angle bisector theorem, <math>BD=\frac{36*40}{36+44} = 18, CD = \frac{44*40}{36+44}=22.</math> Then, by power of a point on <math>D,</math> we have | ||
+ | <cmath>xy=18*22=4*99</cmath> | ||
+ | Subtracting the last two equations, we find | ||
+ | <cmath>x^2=12*99</cmath> | ||
+ | Thus, | ||
+ | <cmath>y^2 = \frac{(xy)^2}{x^2} = \frac{(4*99)^2}{(12*99)} = \boxed{132}</cmath> | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=q0lUoXmkAyk&t=460s |
Latest revision as of 03:40, 25 December 2022
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
.
Solution 3
This is the author's solution. Refer to the above diagrams.
Notice that
So by AA similarity,
It follows that
or
Now let
By the previous result, we have
But by the angle bisector theorem,
Then, by power of a point on
we have
Subtracting the last two equations, we find
Thus,