Comments on: The Facebook Friend http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: .l http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-38302 .l Sat, 25 Sep 2010 07:42:44 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-38302 Haha it's still you coz the video is linked to MissLoi's account. .-= From .l's blog: <a href="http://imeldasanders123.wordpress.com/2010/09/24/hint-19-wow-so-fast/" rel="nofollow">Hint 19 Wow- so fast!</a> =-. Haha it's still you coz the video is linked to MissLoi's account.
.-= From .l's blog: Hint 19 Wow- so fast! =-.

]]> By: Sadako Loi http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30830 Sadako Loi Wed, 14 Oct 2009 17:04:30 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30830 <a href="/miss-loi-the-tutor/revenge-of-the-fallen" rel="nofollow">Too late</a>. Too late.

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30654 Miss Loi Thu, 08 Oct 2009 12:01:48 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30654 *kowtows to <b>mathslover</b> for your never say die attitude* 辛苦你了 ... *kowtows to mathslover for your never say die attitude*

辛苦你了 ...

]]> By: mathslover http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30463 mathslover Fri, 02 Oct 2009 15:20:00 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30463 Knew there was an easier way when I took > 30 mins to solve this supposedly O level question... and [pmath]super*10^infty[/pmath] glad that my real Os are long over. Introducing my ugly handwriting... http://img40.imageshack.us/i/72643930.jpg/ http://img89.imageshack.us/i/72641804.jpg/ http://img134.imageshack.us/i/16454903.jpg/ Last but not least.. this is [pmath]super*10^infty[/pmath] fun! :D Knew there was an easier way when I took > 30 mins to solve this supposedly O level question...

and super*10^infty glad that my real Os are long over.

Introducing my ugly handwriting...
http://img40.imageshack.us/i/72643930.jpg/
http://img89.imageshack.us/i/72641804.jpg/
http://img134.imageshack.us/i/16454903.jpg/

Last but not least.. this is super*10^infty fun! :D

]]> By: mathslover http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30462 mathslover Fri, 02 Oct 2009 14:53:40 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30462 Since when there was intercept theorem?! I went to use similar triangles (and 绕一大圈 to get DM)... And I didn't see the phantom height <i>h</i>! There are 3 similar triangles there (BEM, CEM, CAD) and 3 possible variations of sin x ((EM/BE), (EM/CE), part of area of ABE), and it took me ages to figure out which combination will result in terms cancelling each other! And I even attempted to use double angle (sin 2x = AD/AB = 2 sin x cos x) since it contains the needed line AB. *kowtows* Saw the bit at 0.18 by accident, when pausing it to draw the triangle on paper. But it didn't help me the way Sadako intended... I didn't think that much into it... And saw the other bit after reading your previous post, but the only way I could figure out how to use that information ... is in the lengthy prove. :( Since when there was intercept theorem?! I went to use similar triangles (and 绕一大圈 to get DM)...

And I didn't see the phantom height h!

There are 3 similar triangles there (BEM, CEM, CAD) and 3 possible variations of sin x ((EM/BE), (EM/CE), part of area of ABE), and it took me ages to figure out which combination will result in terms cancelling each other!

And I even attempted to use double angle (sin 2x = AD/AB = 2 sin x cos x) since it contains the needed line AB.

*kowtows*

Saw the bit at 0.18 by accident, when pausing it to draw the triangle on paper. But it didn't help me the way Sadako intended... I didn't think that much into it...

And saw the other bit after reading your previous post, but the only way I could figure out how to use that information ... is in the lengthy prove. :(

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30455 Miss Loi Fri, 02 Oct 2009 13:11:10 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30455 And so if you had watched the video <em>very</em> closely, your subconscious mind will tell you that the key to this supposedly difficult question lies in utilising the <em>areas of triangles</em> (unless someone can think of another approach). To round this off, here's Miss Loi's (similar) approach, after having her mathematical powers similarly triggered by subliminal messages in the video: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/10/sadako-plane-geometry-solution-2a.gif" alt="Miss Loi's Solution Diagram 1" /> <div style="color:blue;"> First we try to obtain an expression for <var>DM</var> on the <acronym title="Left-hand side">LHS</acronym>: As described by <b>mathslover</b> above, Δ<var>BEC</var> is isosceles ⇒ <var>EM</var> ⊥ <var>BC</var> (<var>M</var> mid-pt. of <var>BC</var>) ⇒ <var>EM</var> // <var>AD</var> By Intercept Theorem, [pmath]DM/MC = EA/CE[/pmath] [pmath]DM = MC * {EA/CE}[/pmath] --- (1) </div> <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/10/sadako-plane-geometry-solution-2b.gif" alt="Miss Loi's Solution Diagram 2" /> <div style="color:red;"> Now there are many ways to utilize the areas of the various triangles to obtain an expression that contains <var>AB</var> for the <acronym title="Right-hand side">RHS</acronym> of the proof (see <b>mathslover</b>'s workings for e.g.), Miss Loi shall use the shortest and quickest one here: Comparing the areas of Δ<var>BCE</var> and Δ<var>ABE</var> and expressing them using different formulae: [pmath]{{1/2}AB*BE sin x}/{{1/2} BC*BE sin x}={{1/2}EA*h}/{{1/2}CE*h}[/pmath] <span class="fineprint">(note that they share a common height <var>h</var>)</span> [pmath]AB/BC = EA/CE[/pmath] --- (2) Sub (2) into (1) [pmath]DM = MC * {AB/BC}[/pmath] [pmath]DM = {MC/BC} * AB[/pmath] [pmath]DM = {1/2} * AB[/pmath] (<var>MC</var> = [pmath]1/2[/pmath]<var>BC</var>) <strong class="big">PROVED!!!</strong> *burns a joss stick to Sadako ...* </div> So may Sadako be appeased by our valiant effort, and hopefully all of you can now add a new weapon (i.e. areas of triangles) from beyond your standard <a href="http://www.exampaper.com.sg/blog/wp-content/uploads/geometric-formulae-for-plane-geometry.pdf" rel="nofollow">Plane Geometry handbook</a> to your arsenal to fight that hideous Plane Geometry monster! And so if you had watched the video very closely, your subconscious mind will tell you that the key to this supposedly difficult question lies in utilising the areas of triangles (unless someone can think of another approach).

To round this off, here's Miss Loi's (similar) approach, after having her mathematical powers similarly triggered by subliminal messages in the video:

Miss Loi's Solution Diagram 1

First we try to obtain an expression for DM on the LHS:

As described by mathslover above,

ΔBEC is isosceles
⇒ EM ⊥ BC (M mid-pt. of BC)
⇒ EM // AD

By Intercept Theorem,
DM/MC = EA/CE
DM = MC * {EA/CE} --- (1)

Miss Loi's Solution Diagram 2

Now there are many ways to utilize the areas of the various triangles to obtain an expression that contains AB for the RHS of the proof (see mathslover's workings for e.g.), Miss Loi shall use the shortest and quickest one here:

Comparing the areas of ΔBCE and ΔABE and expressing them using different formulae:
{{1/2}AB*BE sin x}/{{1/2} BC*BE sin x}={{1/2}EA*h}/{{1/2}CE*h} (note that they share a common height h)
AB/BC = EA/CE --- (2)

Sub (2) into (1)
DM = MC * {AB/BC}
DM = {MC/BC} * AB
DM = {1/2} * AB (MC = 1/2BC)

PROVED!!! *burns a joss stick to Sadako ...*

So may Sadako be appeased by our valiant effort, and hopefully all of you can now add a new weapon (i.e. areas of triangles) from beyond your standard Plane Geometry handbook to your arsenal to fight that hideous Plane Geometry monster!

]]> By: mathslover http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30317 mathslover Tue, 29 Sep 2009 18:49:36 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30317 Add on: 2) triangle BEM is similar to triangle CEM (SAS rule) Hope its clearer. And show that 我没有用猜的。 Add on:
2) triangle BEM is similar to triangle CEM (SAS rule)

Hope its clearer. And show that 我没有用猜的。

]]> By: mathslover http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30316 mathslover Tue, 29 Sep 2009 18:38:32 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30316 我知道了！！！ Since BE = EC and BM = MC, we can conclude that: 1) triangle BEC is isosceles angle EBC = angle ECB = x. Hence angle ABE = x since B is 2 times C (given). And since triangle BEC is isosceles, we can conclude that: 2) triangle BEM is similar to triangle CEM Since angle EMB + angle EMC = 180 degrees (straight line), and triangle BEM is similar to triangle CEM, so angle EMB = angle EMC = 90 degrees. 3) And thus we can conclude that triangle CEM is similar to triangle CAD (the AAA rule of similarity). 4) sin x = EM/EB (SOH rule) Ok, proving starts. :P Area of triangle ABC = area of triangle ABE + area of triangle EBC. (1/2)(BC)(AD) = (1/2)(BC)(EM) + (1/2)(AB)(BE)sin x dividing throughout by 1/2, (BC)(AD) = (BC)(EM) + (AB)(BE)sin x substituting sin x = EM/EB, (BC)(AD) = (BC)(EM) + (AB)(BE)(EM/EB) cancelling EB in the last term, (BC)(AD) = (BC)(EM) + (AB)(EM) factorise, (BC)(AD) = (EM)(BC+AB) shifting terms around, (EM/AD) = (BC)/(BC+AB) -- (1) By the similar triangles CEM and CAD (proven earlier), (EM/AD) = (CM/CD) substituting into equation (1), (CM/CD) = (BC)/(BC+AB) shifting terms again, (BC)(CD) = (CM)(BC+AB) opening brackets, (BC)(CD) = (CM)(BC) + (CM)(AB) re-factorising, (BC)(CD-CM) = (CM)(AB) The length of CD - CM is DM (from the diagram given) substituting, (BC)(DM) = (CM)(AB) shifting terms again, DM = (CM)(AB)/(BC) The ratio of CM/BC = 1/2 since M is midpoint of BC. substituting, DM = 1/2 AB SHOWN! Ok Sadako Loi stop cursing me with careless mistakes... :( <div class="highlight"> <b>Miss Loi:</b> That's quite a major surgery you have there but basically you've gotten the approach right i.e. prove that <var>EM</var> is perpendicular to <var>BC</var>, set your base angle <var>x</var> and then obtain an expression using the <em>areas of various triangles</em> to finally get your proof! <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/10/sadako-plane-geometry-solution-1.gif" alt="Mathslover Solution Diagram" /> Did you prove this all by yourself or did you "see" something within that video which somehow subconsciously triggered your inner mathematical prowess? ;) </div> 我知道了！！！

Since BE = EC and BM = MC, we can conclude that:
1) triangle BEC is isosceles

angle EBC = angle ECB = x. Hence angle ABE = x since B is 2 times C (given).

And since triangle BEC is isosceles, we can conclude that:
2) triangle BEM is similar to triangle CEM

Since angle EMB + angle EMC = 180 degrees (straight line), and triangle BEM is similar to triangle CEM, so angle EMB = angle EMC = 90 degrees.

3) And thus we can conclude that triangle CEM is similar to triangle CAD (the AAA rule of similarity).

4) sin x = EM/EB (SOH rule)
Ok, proving starts. :P

Area of triangle ABC = area of triangle ABE + area of triangle EBC.

(1/2)(BC)(AD) = (1/2)(BC)(EM) + (1/2)(AB)(BE)sin x

dividing throughout by 1/2,
(BC)(AD) = (BC)(EM) + (AB)(BE)sin x

substituting sin x = EM/EB,
(BC)(AD) = (BC)(EM) + (AB)(BE)(EM/EB)

cancelling EB in the last term,
(BC)(AD) = (BC)(EM) + (AB)(EM)

factorise,
(BC)(AD) = (EM)(BC+AB)

shifting terms around,
(EM/AD) = (BC)/(BC+AB) -- (1)

By the similar triangles CEM and CAD (proven earlier),
(EM/AD) = (CM/CD)

substituting into equation (1),
(CM/CD) = (BC)/(BC+AB)

shifting terms again,
(BC)(CD) = (CM)(BC+AB)

opening brackets,
(BC)(CD) = (CM)(BC) + (CM)(AB)

re-factorising,
(BC)(CD-CM) = (CM)(AB)

The length of CD - CM is DM (from the diagram given)

substituting,
(BC)(DM) = (CM)(AB)

shifting terms again,
DM = (CM)(AB)/(BC)

The ratio of CM/BC = 1/2 since M is midpoint of BC.

substituting,
DM = 1/2 AB

SHOWN! Ok Sadako Loi stop cursing me with careless mistakes... :(

Miss Loi: That's quite a major surgery you have there but basically you've gotten the approach right i.e. prove that EM is perpendicular to BC, set your base angle x and then obtain an expression using the areas of various triangles to finally get your proof!

Mathslover Solution Diagram

Did you prove this all by yourself or did you "see" something within that video which somehow subconsciously triggered your inner mathematical prowess? ;)

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30311 Miss Loi Tue, 29 Sep 2009 17:01:31 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30311 *Miss Loi's eyes rolls backwards as she looks up suddenly and speaks in a trance-like state* <b>Nash:</b> Sadako's question isn't easy but word from the netherworld has it that <em>subliminal</em> clues have been planted within that video. As for your 42-day quest, Miss Loi's finger begins to move towards this <a href="/miss-loi-the-tutor/within-the-realm-of-a-dying-sun" rel="nofollow">link</a> (which should be relevant for A Levels as well) *Miss Loi's eyes rolls backwards as she looks up suddenly and speaks in a trance-like state*

Nash: Sadako's question isn't easy but word from the netherworld has it that subliminal clues have been planted within that video.

As for your 42-day quest, Miss Loi's finger begins to move towards this link (which should be relevant for A Levels as well)

]]> By: Nash http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30307 Nash Tue, 29 Sep 2009 15:18:44 +0000 http://www.exampaper.com.sg/questions/a-maths/the-facebook-friend#comment-30307 HI miss Loi, just a random person benefiting from your website. Could you post the solutions to sadaoko's question? i cant realate DM to AB somehow. By the way, how can i improve my maths in 42 days before A levels? I've memorised all my formulae but my results arnt improving much. Should i just chiong TYS for the next 1 month? HI miss Loi, just a random person benefiting from your website. Could you post the solutions to sadaoko's question? i cant realate DM to AB somehow. By the way, how can i improve my maths in 42 days before A levels? I've memorised all my formulae but my results arnt improving much. Should i just chiong TYS for the next 1 month?

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