Mathematics Extension 1 Marathon HSC 09 (2 Viewers)

blaze.bluetane · Oct 19, 2009

untouchablecuz said:
i tried yesterday and today, i cant get that goddam upper bound on the angle

its probably something simple

0<=t<1
since theta=half.ln[(1+t)/(1-t)]+pi/4
for any lnx, x>0

blaze.bluetane · Oct 19, 2009

untouchablecuz · Oct 19, 2009

blaze.bluetane said:
0<=t<1
since theta=half.ln[(1+t)/(1-t)]+pi/4
for any lnx, x>0

yep tried that already

all you end up with $\frac{\pi}{4}<x \text { and } 0<2$

jetblack?

lychnobity · Oct 19, 2009

blaze.bluetane said:

i) angle QBX = angle DAC (angles subtending same arc)

angle PAX + angle APQ = angle AXB + angle BXQ (sum of int. angles = opp ext angle in a triangle)

angle PAX = angle BXQ (angle APQ & AXB = 90)
angle PAX is the same as angle DAC, .'. angle QBX = angle BXQ

ii) angle AXP = angle QXC (vert opp angles)

In triangles APX & BXC:
angle APX = angle BXC (both =90)
angle PAX = angle XBC (angle subtending same arc)
angle AXP = angle BCX (3rd angle)
.'. triangle PAX is similar to triangle BXC, angle AXP = angle BXC (corresp. angles in similar triangles)

.'. angle AXP = angle QXC = angle BCX

.'. triangle QXC is isos, QX = QC

But QX = QB (triangle QBX is isos, from above)

.'. QB = QC, Q bisects BC

EDIT: sorry couldn't think of a quicker method.

untouchablecuz · Oct 19, 2009

does

not

compute

lychnobity · Oct 19, 2009

untouchablecuz said:
does

not

compute

...i tried

ninetypercent · Oct 19, 2009

blaze.bluetane said:
scardizzle +1

nice handwriting

blaze.bluetane · Oct 19, 2009

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lychnobity said:
i) angle QBX = angle DAC (angles subtending same arc)

angle PAX + angle APQ = angle AXB + angle BXQ (sum of int. angles = opp ext angle in a triangle)

angle PAX = angle BXQ (angle APQ & AXB = 90)
angle PAX is the same as angle DAC, .'. angle QBX = angle BXQ

ii) angle AXP = angle QXC (vert opp angles)

In triangles APX & BXC:
angle APX = angle BXC (both =90)
angle PAX = angle XBC (angle subtending same arc)
angle AXP = angle BCX (3rd angle)
.'. triangle PAX is similar to triangle BXC, angle AXP = angle BXC (corresp. angles in similar triangles)

.'. angle AXP = angle QXC = angle BCX

.'. triangle QXC is isos, QX = QC

But QX = QB (triangle QBX is isos, from above)

.'. QB = QC, Q bisects BC

EDIT: sorry couldn't think of a quicker method.

nice. as always there are alternative methods. my method was shorter for part ii but longer for part i.
quicker method for part 2...
(I always try to start part ii with the statement 'using part i')
1. QBX is isosceles (2 equal angles, from part i), QB=QX
2. Similar to part i, angleQXC=angleQCX
(just rotate the diagram 180 and a similar situation is presented)
3. As you said, QXC is isosceles
4. QB=QX=QC, Q bisects BC

lychnobity's turn

lychnobity · Oct 20, 2009

blaze.bluetane said:
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nice. as always there are alternative methods. my method was shorter for part ii but longer for part i.
quicker method for part 2...
(I always try to start part ii with the statement 'using part i')
1. QBX is isosceles (2 equal angles, from part i), QB=QX
2. Similar to part i, angleQXC=angleQCX
(just rotate the diagram 180 and a similar situation is presented)
3. As you said, QXC is isosceles
4. QB=QX=QC, Q bisects BC

lychnobity's turn

Wow, nice

Alright, here's another circle geometry one

lychnobity · Oct 20, 2009

jetblack2007 said:
After hermand's question is done:

$\text{As a particle moves around a circle, its angular velocity is given by } \\\frac{d\theta}{dt} = \frac{1}{1 - t^2} \\ \text{You are given that the particle starts at } \theta = \frac{\pi}{4} \\ \text{Find t as a function of }\theta \text{ and prove that } \frac{\pi}{4} \leq \theta < \frac{3\pi}{4} \text{ and explain why the particle }\\\\\text{ never moves through an angle of more than } \frac{\pi}{2}$

$\int {\mathrm{d\theta}} = \int\frac{{\mathrm{d} t}}{1-t^{2}}\\\\\theta = \int\frac{\frac{1}{2}}{1-t} + \frac{\frac{1}{2}}{1+t}\ \mathrm{dt} = \frac{1}{2}\ln \left(\frac{1+t}{1-t}\right ) + C\\\\ \theta = \frac{\pi}{4}, t = 0\\\frac{\pi}{4} = \ln\left ( 1 \right ) + C \Rightarrow C = \frac{\pi}{4}\\\\\therefore \theta = \frac{1}{2}\ln\left(\frac{1+t}{1-t} \right ) + \frac{\pi}{4}\\\\\frac{1+t}{1-t} = e^{2\theta - \frac{\pi}{2}}\\\\ \text{Mutiply both sides by (1-t) then minus 1}\\\\ t = e^{2\theta - \frac{\pi}{2}}(1-t) -1\\\\ t(1 + e^{2\theta - \frac{\pi}{2}}) = e^{2\theta - \frac{\pi}{2}} -1\\\\ t = \frac{e^{2\theta - \frac{\pi}{2}}-1}{1 + e^{2\theta - \frac{\pi}{2}}}$

To jetblack, is the question meant to have a plus sign, instead of a minus? I had to use partial fractions, and I can't do the second or third parts.

untouchablecuz · Oct 20, 2009

lychnobity said:
Wow, nice

Alright, here's another circle geometry one

a)

AOM=90 (given)

APM=180-APB=90 (angles on a straight line)

.'. since AOM=APM=90, AOPM is cyclic (if two angles subtended by the same two points are equal, then the points lie on a unique circle)

b)

OMB=OAP (angles subtended by the same arc are equal) [1]

AO=OP (equal radii) --> triangle AOP is isoceles

OAP=OPA (base angles of isoc. triangle) [2]

from [1] and [2], OMB=OPA

blaze.bluetane · Oct 20, 2009

untouchablecuz said:
a)

AOM=90 (given)

APM=180-APB=90 (angles on a straight line)

.'. since AOM=APM=90, AOPM is cyclic (if two angles subtended by the same two points are equal, then the points lie on a unique circle)

b)

OMB=OAP (angles subtended by the same arc are equal) [1]

AO=OP (equal radii) --> triangle AOP is isoceles

OAP=OPA (base angles of isoc. triangle) [2]

from [1] and [2], OMB=OPA

very smooth.

But I think you need to say in part 1 that angleAPB = 90 BECAUSE it is the angle at the circumference of a semicircle. I don't think angleAPB=90 is a given.

scardizzle · Oct 20, 2009

Regarding that question jetblack asked, that's 4U intergration right?
I've never come across an intergration question that requires you to split the denominator in a 3U paper... have i just been living under a rock?

untouchablecuz · Oct 20, 2009

blaze.bluetane said:
very smooth.

But I think you need to say in part 1 that angleAPB = 90 BECAUSE it is the angle at the circumference of a semicircle. I don't think angleAPB=90 is a given.

ah yes

thanks son

Trebla · Oct 20, 2009

Solve for x:
ln(x + 2e) + ln(x - 2e) = 1 + ln 3x

lychnobity · Oct 21, 2009

Trebla said:
Solve for x:
ln(x + 2e) + ln(x - 2e) = 1 + ln 3x

$\ln\left ( \frac{x^{2} - 4e^{2}}{3x} \right )=1\\\\\frac{x^2 - 4e^2}{3x} = e\\\\ x^2 -4e^2 = 3ex\\(x+e)(x-4e)= 0 \\\therefore x = -e \or \ x = 4e$

Timothy.Siu · Oct 21, 2009

lychnobity said:
$\ln\left ( \frac{x^{2} - 4e^{2}}{3x} \right )=1\\\\\frac{x^2 - 4e^2}{3x} = e\\\\ x^2 -4e^2 = 3ex\\(x+e)(x-4e)= 0 \\\therefore x = -e \or \ x = 4e$

are u sure x can = -e

Trebla · Oct 21, 2009

Yep,

There is clearly a restriction on the log fuctions:
ln(x + 2e) + ln(x - 2e) = 1 + ln 3x

For the equation to be valid:
- x + e > 0
- x - 2e > 0
- 3x > 0

The common inequality of the three restrictions is x > 2e, hence x = - e is not a defined solution (so only x = 4e is valid)

jet · Oct 21, 2009

Hahahahahah, guys I'm so sorry, that motion question is meant to have a plus like Lychnobility asked. I'm so sorry lol. Sometimes my brain just doesn't compute. Now it should be solvable.

blaze.bluetane · Oct 21, 2009

ok well in that case,

Fair bit easier than splitting denominators and log results and all that!

Mathematics Extension 1 Marathon HSC 09 (2 Viewers)

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