Question

In the attached picture, the angle \(\theta\) is constant. CA, DE and CB are tangent to the circle.
M is the midpoint of the arc AB. Compute in terms of
\(\theta\) the ratio of similarity of the triangles CAB and CDE.

Answer 1

Should this be done with coordinate geometry?

Answer 2

You can do it with basic trigonometry.

Answer 3

Trial..... Probably wrong.....
Join DO and MC.
Let AO = r
AO = BO = MO = r (radii)

Angle ACO = θ/2 (prop. Of tangents)
Triangle OCA~ triangle OAN
Angle OAB =θ/2 (corr. Angles, ~ triangles)

AN = r cos(θ/2)
AB = 2r cos(θ/2)

Angle AOB = 180 – 2(θ/2) = 180-θ (angle sum of triangle)
Angle AOM = 90-(θ/2)
Angle DOM = 45 –(θ/4)

DM = r tan [45-(θ/4)]
DE = DM = 2r tan [45-(θ/4)]
Ratio of similarity of triangle CAB and CDE
= [2r cos(θ/2)] / [2r tan [45-(θ/4)] ]
= cos(θ/2) / tan [45-(θ/4)]

Doesn't look good at all :|

Answer 4

Where's point N? "Triangle OCA~ triangle OAN"

Answer 5

Oh right I saw the diagram

Answer 6

The answer's probably really easy and we've all missed it XD

Answer 7

Yup... it's not that ugly as what I've worked out for now, I believe :(
Any ideas?!

Answer 8

I've forgotten the identities, could it be possible that angle AMB is 2*ACB?

Answer 9

angle AOB = 180 - θ
reflex angle AOB = 360 - (180 - θ) = 180 + θ
Angle AMB = ( 180+θ) /2 = 90 + θ/2 (angle at centre, twice angle at circumference)

Answer 10

and angle ACB = θ (given)

Answer 11

But in terms of θ, I don't expect any other unknowns?!

Answer 12

Made a few errors...

Answer 13

Coming soon...

Answer 14

Yes in terms of \(\theta \) only.

Answer 15

Dr, no trigo ratio involved in the final answer?!

Answer 16

May be, but you can simplify into one trig function.

Answer 17

surely there has to be a trig ratio in the final answer, or else what would you do with the angle?

Answer 18

Hmm.. ''may be'' That means the answer he's got has no trigo ratio :|

Answer 19

The answer I got should looks like
\[
K \cos^2( u(\theta))
\]
Where u is a function of \(\theta \)

Answer 20

I have a cos^2 !!!

Answer 21

K is a numerical constant?

Answer 22

K is a numerical constant.

Answer 23

Here's what I got

angle ANO = 90°, as is angle OAC
r = AO, r = MN + ON
∆ANO ||| ∆CAO

angle OAN = 90-θ/2
sin OAN = ON/AO
ON = r sin(90-θ/2) = r cos(θ/2)

Now to ∆ CAN, which is ||| ∆AON
angle CAN = 90-θ/2
tan(90-θ/2) = (NM + CM)/AN
= (r-a + CM)/AN
lets say x = MC
tan (90-θ/2) = cot (θ/2) = (r-a+x)/r cos(θ/2)
r*cos^2(θ/2)/sin(θ/2) = r-a+x
x = r*cos^2(θ/2)/sin(θ/2) - r + a
x/(x+r-a) should be your ratio.

Now to put it in the format you want...

Answer 24

x ... r .... a.... = unknowns!

Answer 25

Yeah, that's the problem, but at least its a start

Answer 26

x = r*cos^2(θ/2)/sin(θ/2) - r + a

Answer 27

a can be easily found

Answer 28

so that just leaves the r

Answer 29

a = which side?

Answer 30

The answer is independent of r.

Answer 31

so r doesn't matter... (oh yeah, it shouldn't matter for similar figures)

Answer 32

@Callisto a = ON

Answer 33

Dr, was I on the right track in my trial?

Answer 34

Believe me it does not depend on r.

Answer 35

Dr..... I didn't include r in my answer.... Just.. some not good-looking trigo ratio....

Answer 36

I gotta go now... :D

Answer 37

Work in the two triangles ODM and OAN. See the attached.

Answer 38

@Callisto, how does you answer look like?

Answer 39

= cos(θ/2) / tan [45-(θ/4)]
= cos(θ/2) : tan [45-(θ/4)]
I've been working on these two triangles...

Answer 40

What's \( \sin (\alpha) \) and \( \tan (\alpha/2) \) in the two triangles
AON and ADN?

Answer 41

Sir... Are you sure you're asking for triangle ADN?

Answer 42

@Callisto, I think your answer is right. Sorry ADM.

Answer 43

Do you mean ODM?

Answer 44

AON and ODM (sorry for that)

Answer 45

It's okay :)

In triangle AON
sin α = AN / AO

In triangle ODM
tan (α/2) = DM/OM

Answer 46

Note that OA = OM (radii)

Answer 47

\[
\sin(\alpha)=\frac {AN} R\\
\tan(\alpha/2)=\frac {DM} R\\
\frac {AN}{DM}= \frac {\sin(\alpha)} {\tan (\alpha/2}=
\frac { 2 \sin(\alpha/2) \cos(\alpha/2)}
{\frac {\sin(\alpha/2)}{\cos(\alpha/2)} }=\\
2 \cos^2 (\alpha/2) =2 \cos^2 (\frac \pi 4 -\frac \theta 4)

\]

Answer 48

Wow! Amazing!!!!

Answer 49

@Callisto, as I said before your answer in term of tan and cos^2 is right.

Answer 50

No, sir... My answer was in cos only, not cos^2.
They don't seem to be equal :(

Answer 51

Little amendment to my first post:
DE = 2DM = 2r tan [45-(θ/4)]

I still can't work out if my answer is equal to Dr's answer though :|

Answer 52

\[
\cos(\theta/2) = \sin(\alpha)=2\sin((\alpha/2)\cos((\alpha/2)\\
\tan (\pi/4 -\theta/4) = \tan(\alpha/2)=\frac{ \sin(\alpha/2) }{\cos(\alpha/2)}
\\\frac {\cos(θ/2)}{tan(π/4−θ/4)}=\frac {2\sin((\alpha/2)\cos((\alpha/2)}{\frac{ \sin(\alpha/2) }{\cos(\alpha/2)}}= 2 \cos^2(\alpha/2)=\\
2 \cos^2(\pi/4 -\theta/4)

\]
@Callisto your answer is the same as mine.

Answer 53

Oh... Thanks Dr!!!