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Let $AC$ be a diameter of a circle $\omega$ of radius $1$, and let $D$ be the point on $AC$ such that $CD = \frac{1}{5}$. Let $B$ be the point on $\omega$ such that $DB$ is perpendicular to $AC$, and let $E$ be the midpoint of $DB$. Compute length $AE$.

 Mar 8, 2024
 #1
avatar+129690 
+2

AO = 1  = CO

AD  =   AO + CO  - CD  =  1 + 1  - 1/5  =   2 -1/5 = 9/5

DC = 1/5

AC perpendicular to BF  ....so  chord BF is bisected by AC

So.....BD = DF

Intersecting Chord Theorem

AD * DC =  BD * DF

AD * DC  = BD * BD

AD * DC = BD^2

(9/5) (1/5) = BD^2

9/25 = BD^2

BD = 3/5

DE = (1/2)(BD)   = (1/2)(3/5)  = 3/10

 

Triangle AED is right

 

Pythagorean Theorem

 

AE^2 = AD^2 + DE^2

 

AE^2  = (9/5)^2 + (3/10)^2

 

AE^2  = 81/25 + 9/100

 

AE^2  =  [324 + 9 ] / 100

 

AE^2  = 333/100

 

AE = sqrt (333) / 10 ≈ 1.82

 

 

cool cool cool

 Mar 8, 2024
edited by CPhill  Mar 8, 2024

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