Power of a Point Theorem/Introductory Problem 3: Difference between revisions
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Letting <math> AR = x </math> makes <math> BR = 4x </math>. Similarly, letting <math> CR = 4y </math> makes <math> DR = 9y </math>. Thus <math> AB = AR + BR = x + 4x = 5x </math> and <math> CD = CR + DR = 4y + 9y = 13y </math>. We therefore seek <math> \frac{AB}{CD} = \frac{5x}{13y} </math>. | Letting <math> AR = x </math> makes <math> BR = 4x </math>. Similarly, letting <math> CR = 4y </math> makes <math> DR = 9y </math>. Thus <math> AB = AR + BR = x + 4x = 5x </math> and <math> CD = CR + DR = 4y + 9y = 13y </math>. We therefore seek <math> \frac{AB}{CD} = \frac{5x}{13y} </math>. | ||
From the Power of a Point Theorem we have that | From the Power of a Point Theorem, we have that | ||
<center><math> x\cdot 4x = 4y\cdot 9y\Rightarrow \left(\frac xy\right)^2 = 9 </math></center> | <center><math> x\cdot 4x = 4y\cdot 9y\Rightarrow \left(\frac xy\right)^2 = 9 </math>,</center> | ||
which gives <math> \frac xy = \pm 3 </math> so we take <math> \frac xy = 3 </math>. | which gives <math> \frac xy = \pm 3 </math>, so we take <math> \frac xy = 3 </math>. | ||
Finally | Finally, | ||
<center><math> \frac{5x}{13y}=\frac 5{13}\cdot \frac xy = \frac 5{13}\cdot 3 = \frac{15}{13}. </math></center> | <center><math> \frac{5x}{13y}=\frac 5{13}\cdot \frac xy = \frac 5{13}\cdot 3 = \frac{15}{13}. </math></center> | ||
''Back to the [[Power of a Point Theorem]].'' | ''Back to the [[Power of a Point Theorem]].'' | ||
Latest revision as of 13:40, 3 July 2006
Problem
(ARML) In a circle, chords 
and 
intersect at 
. If 
and 
, find the ratio 
Solution
Letting 
makes 
. Similarly, letting 
makes 
. Thus 
and 
. We therefore seek 
.
From the Power of a Point Theorem, we have that
,which gives 
, so we take 
.
Finally,
Back to the Power of a Point Theorem.
