Extension One Revising Game (1 Viewer)

[clintmyster]

it requires us to scan diagrams which is the gay thing!

 

[Trebla]

Not exactly "difficult" but...

Consider an isosceles triangle ABC with AB = AC with equal angles /_ABC = /_ACB = 80°.
A line is drawn from point B to point E on AC such that /_CBE = 20°.
Show that points A, B and E lie on a circle with BC as a tangent to this circle.

 

[kurt.physics]

Not exactly "difficult" but...

Consider an isosceles triangle ABC with AB = AC with equal angles /_ABC = /_ACB = 80°.
A line is drawn from point B to point E on AC such that /_CBE = 20°.
Show that points A, B and E are concyclic.

Trebla,

Isnt this trivial, "Any 3 non-collinear points are concyclic"?

 

[Trebla]

Edited

 

[kurt.physics]

Edited

where was it edited? If it is, then the same triviality applys?

edit: sorry, it just showed up :S

 

[kurt.physics]

Not exactly "difficult" but...

Consider an isosceles triangle ABC with AB = AC with equal angles /_ABC = /_ACB = 80°.
A line is drawn from point B to point E on AC such that /_CBE = 20°.
Show that points A, B and E lie on a circle with BC as a tangent to this circle.

Its trivial that 3 non-collinear points are concyclic,

hence points A, B and E lie on a circle

For BC to be tangent, it is neccesary to show that the angle in the alternate segement is equal to the angle between the tangent and chord. ie we must show /_CBE = /_BAE.

/_BEC + 20 + 80 = 180 (/_sum triangle BEC)

/_BEC = 80

/_AEB = 180 - 80 (adj sup /_'s)
= 100

Also,

/_ABF + 20 = 80 (adj /_,s)

/_ABF = 60,

/_BAE + 60 + 100 = 180 (/_sum triangle ABE)

/_BAE = 20

Its given that /_CBE = 20,

hence

/_CBE = /_BAE

thus BC is tangent to this circle.

QED

 

[kurt.physics]

I will leave this difficult circle geometry

AB is a diameter of a circle, PQ is a chord of the circle, perpendicular to AB and nearer to B, cutting AB at V. M is any point on AV. QM produced cuts the circle at R. Show that /_RPA = /_MPA.

 

[bored of sc]

think arg (z^n)=narg(z), and umm.. arg(z^n)=kpi, where K is 0, +-1,+-2,...

Oh!
You want the argument of zⁿ to be kpi so the sin@ = 0 and thus 'i' disappears.

arg(z) = 75 and arg(zⁿ) = kpi = narg(z) = 75n

So, 75n = kpi where k is any integer.

Now is it just guess and check?

k not 0 since n = 0, which isn't a positive integer
k not less than 0 since n < 0, which isn't a positive integer
k not less 5 since n is not an integer
k = 5 i.e. 75n = 5*180 = 900
n = 900/75 = 12

n = 12 is the smallest positive integer which makes the equation true

z¹² = [rt2/2*cis75]¹²
= 64/4096 * cis900
= 1/64 * -1
= -1/64

*crosses fingers*

 

[5647382910]

I will leave this difficult circle geometry

AB is a diameter of a circle, PQ is a chord of the circle, perpendicular to AB and nearer to B, cutting AB at V. M is any point on AV. QM produced cuts the circle at R. Show that /_RPA = /_MPA.

Draw PR, PA and PM in and let centre be O
since OB is perp. to PQ, PV = VQ (perpendicular from centre to chord bisects chord)
now, angle AVP = angle AVQ = 90 degrees. PV = VQ and MV is common to triangles MVP and MVQ. Therefore triangle MVP is congruent to triangle MVQ (SAS) . therefore MQ = MP and angle QMV = angle VMP.
Now, draw AQ. Angle AMQ = angle AMP as angle QMV = angle VMP.
let angle RPA be x. Therefore angle AQM = x (angle in same segment standing on arc AR
looking at triangles APM and AMQ; AM is common, MQ = MP (from earlier) and angle AMQ = angle AMP (from earlier). therefore triangles AMP and AMQ are congruent (SAS). Therefore angle MPA = angle AQM = x
therefore angle RPA = angle MPA

TO continue the trend....
Two circles (one large one small) touch internally at A. The tangent at a point P on the smaller circle cuts the larger circle at Q and R. Prove that AP bisects angle RAQ

 

[bored of sc]

I tried doing the above question and falied.

Here's another from Fitzpatrick:

Two chords AB and CD of a circle meet when produced at a point P outside the circle. Prove that triangle ADP and triangle CBP are similar.

 

[5647382910]

I tried doing the above question and falied.

Here's another from Fitzpatrick:

Two chords AB and CD of a circle meet when produced at a point P outside the circle. Prove that triangle ADP and triangle CBP are similar.

LOL, ur not following the rules, you gotta answer my Q first;
ill give u a hint,
if you produce the tangent at P and A to meet at a point Z, angle PZA = angle AZP.

 

[cyl123]

Another question:
solve for x; (x+1)(x+2)(x+3)(x+4) = 8

Since no one else bothered...

Sub u=x+2.5

So (u-1.5)(u-0.5)(u+0.5)(u+1.5) = 8
So (u^2 - 2.25)(u^2 - 0.25) = 8
So u^4 - 2.5u^2 -7.4375 =0
Using quad eq gives

u^2 = 4.25 or -1.75

Thus (assuming you are allowed complex answers)
u= sqrt(4.25), -sqrt(4.25), sqrt(1.75)*i, -sqrt(1.75)*i
x= -2.5+sqrt(4.25), -2.5-sqrt(4.25), -2.5+sqrt(1.75)*i, -2.5-sqrt(1.75)*i

 

[kurt.physics]

Two chords AB and CD of a circle meet when produced at a point P outside the circle. Prove that triangle ADP and triangle CBP are similar.

This one is fairly easy

Join A to D and C to B

/_BAD = /_BCD (/_'s in same segment)

/_P is common

/_ADP = /_CBP (/_sum triangle)

By AAA,

triangle ADP and CBP are similar.

Now lets go back to 5647382910's problem!<!-- google_ad_section_end --> <script type="text/javascript"> vbmenu_register("postmenu_4034175", true); </script> <!-- google_ad_section_end --> <script type="text/javascript"> vbmenu_register("postmenu_4034175", true) 655 </script>

 

[addikaye03]

Oh!
You want the argument of zⁿ to be kpi so the sin@ = 0 and thus 'i' disappears.

arg(z) = 75 and arg(zⁿ) = kpi = narg(z) = 75n

So, 75n = kpi where k is any integer.

Now is it just guess and check?

k not 0 since n = 0, which isn't a positive integer
k not less than 0 since n < 0, which isn't a positive integer
k not less 5 since n is not an integer
k = 5 i.e. 75n = 5*180 = 900
n = 900/75 = 12

n = 12 is the smallest positive integer which makes the equation true

z¹² = [rt2/2*cis75]¹²
= 64/4096 * cis900
= 1/64 * -1
= -1/64

*crosses fingers*

haha CORRECT! Well done man. Glad to see you get there!

 

[lolman12567]

Another question:
solve for x; (x+1)(x+2)(x+3)(x+4) = 8

(x+1)(x+4)(x+2)(x+3) = 8

(x²+5x+4)(x²+5x+6)=8

let u = x²+5x

(u+4)(u+6)=8

u²+10u+16 = 0

(u+8)(u+2) = 0

u = -2, -8

x²+5x = -2

(quad formula) x= (-5 + sqrt17)/2, (-5 - squrt 17)/2

x²+5x = -8

(quad formula) doesnt work

.'. x= (-5 + sqrt17)/2, (-5 - squrt 17)/2

not sure if this is right tho

 

[bored of sc]

Two circles (one large one small) touch internally at A. The tangent at a point P on the smaller circle cuts the larger circle at Q and R. Prove that AP bisects angle RAQ

Create tangents at A and Q, labelling their intersection Y.
Extend the tangent at P past R, label the intersection with other tangent (at A) Z.
QY = AY (tangents to exterior point equal, larger circle)
PZ = AZ (similarly, for the smaller circle)
hence triangle YQA and ZPA are isosceles (pair of equal sides each) i.e. angle YQA = angle YAQ = $ and angle ZAP = angle ZPA = @

Let the intersection of AQ at smaller circle S, join SP.
Now angle YAQ = angle SPA = $ (smaller circle, angle between chord/tangent = angle in alternate segment)
similarly angle ZAP = angle PSA = @ (smaller circle)
& angle YQA = angle PRA = $ (larger circle)

Therefore angle PSA = angle ZPA = @ and angle SPA = angle PRA = $
Thus triangle PSA is similar to triangle PRA (equiangular)
So angle PAS = angle PAR = 180-(@+$)
Therefore AP bisects angle RAQ

Please be right.

 

[bored of sc]

New question:

(i) Factorise a³+b³

(ii) Hence, or otherwise, show that

(sin³@+cos³@)/(sin@+cos@) = (2-sin2@)/2

 

[lolman12567]

New question:

(i) Factorise a³+b³

(ii) Hence, or otherwise, show that

(sin³@+cos³@)/(sin@+cos@) = (2-sin2@)/2

i) (a+b)(a²-ab+b²)

ii) RHS= 2-2sin@cos@/2
= 1-sin@cos@
LHS=(sin@+cos@)(sin²@-sin@cos@+cos²@)/sin@+cos@
= sin²@-sin@cos@+cos²@
= 1-sin@cos@
LHS=RHS
.'. (sin³@+cos³@)/(sin@+cos@) = (2-sin2@)/2

 

[kurt.physics]

i) (a+b)(a²-ab+b²)

ii) RHS= 2-2sin@cos@/2
= 1-sin@cos@
LHS=(sin@+cos@)(sin²@-sin@cos@+cos²@)/sin@+cos@
= sin²@-sin@cos@+cos²@
= 1-sin@cos@
LHS=RHS
.'. (sin³@+cos³@)/(sin@+cos@) = (2-sin2@)/2

Its your turn to leave a question

 

[lolman12567]

differentiate x³ using first principles.