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#1 Re: Help Me ! » Quadrilateral midpoint vector proof » Yesterday 21:46:56

Freiza wrote:

3) You were born on a Monday

I think you're confusing me with Solomon Grundy   smile

But why MN is not relative to origin, ie why it is not (a+d)/2 + (a+b)/2, i know this is not correct but I am failing the intuition here.

The method works by finding the equation of the two lines and then setting them equal to find the point of intersection. 

To construct such an equation:

Step (1) Start at the origin.
Step (2) Write a vector from the origin to anywhere on the line.
Step (3) Go an unknown amount in the direction of the line using a parameter.

One of the lines goes through the origin so step (2) is not required.

In coordinate geometry it is usual to use x for the independent variable and y for the dependent variable.

In vector geometry a Greek letter and a vector 'r' are usually used.

#2 Re: Help Me ! » Hard Proof » Yesterday 21:42:26

hi ET ag

My method is similar.  I called the points (a,1/a); (b,1/b); and (c,1/c) and worked out the equation of the line perpendicular to AB through C.  Once you have the form for that line, you can write down a similar equation for the line perpendicular to BC through A.

Find where they intersect (most of the algebra cancels out) and show the resulting point is on the hyperbola.  [This also shows that the third altitude goes through the same point.]


#3 Re: Help Me ! » Need some help with transformations problems » Yesterday 21:38:01

hi ET ag

Welcome to the forum.

The forum LaTex doesn't support your draw commands so you won't get an image that way.  Most members use Geogebra to construct their diagrams; save them to an on-line site; and then link from there using bcCode.  A recent post has addressed this.


#4 Re: Help Me ! » Quadrilateral midpoint vector proof » 2017-02-21 23:16:51

hi Freiza,

Ok, that looks better now.

But the intersection has to be relative to the origin:



#5 Re: Help Me ! » Geometry Proof. » 2017-02-21 20:48:51

hi Maroon900

Welcome to the forum.

At first I thought this was a use of the triangle inequality and, after a bit of manipulating, I got AD < AB + AC.  Nearly there I thought.  Then I tried a fresh approach:

Rotate the triangle about point D to create a new triangle A'BC below the first.  The resulting shape ABA'C is a parallelogram (can you prove this?) and the diagonal is 2AD.  You'll get the required result from this.


#6 Re: Help Me ! » 3D Geometry Help Please » 2017-02-21 20:38:19

hi adpqadpq

You could call the size of the base 'x' and form an inequality for x squared using this information.  So that would answer the question if we are meant to dis-regard  any cut off wastage. 

But if we have to start with a sheet of cardboard and design a net for the box with minimum wastage this will be harder to determine.  Please clarify.


#7 Re: Help Me ! » Quadrilateral midpoint vector proof » 2017-02-21 06:20:07

hi Freiza,

Welcome to the forum.

I have embedded your image into your post so that members will not have to go to that site to view it.

Your method should work but it looks to me like there are many errors.

Let's try to sort them out one at a time.  I'll leave out the vector over-lining to save time.

You want the vector for QN.

I would work like this:

QN = QM + MN = QA + AM + MB + BN = d/2 + a/2 + a/2 + b/2 = (d + 2a + b)/2

Are you able to work out PM in a similar way?  Please post back.


#8 Re: Maths Is Fun - Suggestions and Comments » Image upload » 2017-02-21 00:25:49

hi chen.aavaz

Welcome to the forum.

After an update to the forum some time ago it was no longer possible to upload directly in a post.

So now I store my images on (there are other sites where you can do something similar) and then link to it using bc code.  Here's a set if instructions I made earlier: … 06#p352606

post #1686


#9 Re: Help Me ! » Bag with balls » 2017-02-21 00:15:30

hi iasonidis

Yes, you are correct.  I got lazy as it's 1/2 on both choices so it doesn't effect the final answer.


#10 Re: Help Me ! » Bag with balls » 2017-02-19 22:09:51

hi samuel.bradley.99

I think you can do this using Bayes theorem but, as I can never be bothered to learn it, I prefer to make a probability tree diagram.


Working from left to right, we have the initial set up of 10 blacks and x white.  Then either a black or white is added shown by the two branches.

Then we choose a ball which may be black or white.  So a further two branches are shown for each of the branches already added.

The final outcomes are shown as 'its black' given 'black' and so on.

As we know the final ball is black we only have two branches to consider: P(b|b) and P(b|w).  So I think the answer is

Hope that helps,


#11 Re: Help Me ! » Please find the average number » 2017-02-17 23:31:27

hi Alex6969

Welcome to the forum. 

There's a formula for expected value which goes like this:

where the x's  are the values and the p's are the probabilities.  So you could take the x values using those midpoints and the probabilities by dividing the percentages by 100 and adding them together like this:

Some might argue that the midpoint is not 24.5 etc as you ought to be considering a continuum from 0 to 50 inclusive which would make the midpoint 25.  In practice it doesn't make much difference as this is only a theoretical statistic.

You'll also have to make a judgement about what to use for A9.


#12 Re: Help Me ! » Geometry Help Please » 2017-02-17 23:18:15

hi adpqadpq

Welcome to the forum.

There's a difference here between the engineering problem where you actually want to get to the centre without getting your feet wet and the geometry problem where we assume that lines have zero 'thickness'.  The dotted lines on my diagram show that you'd have to overlap the second plank on the first in order to make a secure platform.  But I'll assume we meant to ignore that and take CD as 10 and the chord AB as 16.


It's a property of all circles that a line from the centre of the circle to the centre of any chord will be at right angles to the chord.  In other words angle ADC is 90.  So you can use Pythagoras theorem to calculate AC squared.  Don't bother to find the square root because you want radius squared for the area.  (pi r squared)

Hopefully that should be enough for you to complete the question.


#13 Re: Help Me ! » Toothpick problem: » 2017-02-16 20:17:41

hi Mathegocart

Your method will work but it's not 31 + 9 x 21 but rather 31 + 19 x 21= 31 + 399 = 430.  There are 19 more rows to add.

bobbym.  This is a grid made up of 10 by 20 squares.  Each square has four 'toothpicks'.


#14 Re: Help Me ! » Vectors Question - Direction and Speed » 2017-02-13 20:55:16

hi Oran2009


I have shown the resultant vector (DH) with a double arrow.  It is made up from two components, DP and PH.  You know DH and PH so you can use trig. to get DP.


#15 Re: Help Me ! » Triangle Proofs » 2017-02-03 23:38:45

hi Shrimpy

Welcome to the forum.

It's a while since I did this question but I seems to remember that you can get cos36 from another right angled triangle in the diagram.  Make a sketch and split a 36-36-108 triangle down the middle.

If you're still not getting there post back and I'll start the question again.


#16 Re: Help Me ! » Conic Questions » 2017-01-29 07:45:12

hi RandomMathUsername

If you picked a name at random, how unlikely is your username?  Hhhmmm. smile  But welcome anyway!

I'm still thinking about Q1.

Here's my initial thoughts on Q2.


As those points are on the circle, the perpendicular bisector goes through the centre.  No other constraint so there are an infinite number of centres and circles.

My diagram shows two.  The products (only approximately here) seem to be the same, so I'm conjecturing the product is a constant whatever the circle.

So you could find the equation of the bisector, and obtain an expression for the radius and hence find the coordinates of the other two points.  Use the equation of the line to eliminate y (or x if you prefer) make the product, and hope that x (or y) cancels out of the result.

Good luck!


#17 Re: Help Me ! » Nets of Cubes » 2017-01-23 20:06:40

hi Chelsianna

Welcome to the forum.

Here's one net of a cuboid.


For my example I chose a cuboid 4 by 3 by 2.

You need six rectangles, two are 4 by 3 and they will become the top and bottom of the solid.  Two are 4 by 2.  They will form the back and front faces.  The final two are 3 by 2. They will make the left and right faces.

There are a number of ways that those six rectangles could be joined, and still you'd get a cuboid.  For example you could leave out face B from the right hand end and have a new B on top of E.  That would still work.

But it wouldn't work if you joined A to B directly because there has always got to be a side face between the top and the bottom.

A good way of practising this would be to get some paper with squares and try cutting out different possible nets to see how many ways you an find that work.

A cube is just a cuboid with all sides the same length, which makes it simpler.  You'd need 6 squares that fold up to make a cube.  There are a number of ways to join them, to make this work.  If bobbym says there are 11 ways I'm sure he is correct.  I haven't tried to find them myself.  And I don't think you would be asked this in an exam because it's too open ended for a test against the clock.

But I have seen GCSE papers where three different configurations are given and you're asked to say which will work and which not.  For example if you put all six squares in a line, then this would fold up so some faces came on top of others and some faces would be left out.

You might be given a partly completed net and asked to complete it.  Trying some nets and cutting them out is a good way to get better at this.

What exam is it?  It sounds a bit like GCSE (UK) but maybe it isn't.  If you post back I'll look up and see what the syllabus says and if there are any practice questions.

Hope that helps,


#18 Re: Help Me ! » Geometry: Consecutive sides that are congruent not a paralellogram? » 2017-01-18 23:44:30

If a quadrilateral has one pair of sides that are equal in length and parallel then it is a parallelogram whatever the other sides are 'doing'.


The sides cannot be adjacent as they are parallel so they must be opposite.

Let's suppose the quadrilateral is ABCD lettered in order around the shape and that AB = DC and AB // DC.

Join AC.

AC is a transversal so angle BAC = DCA .

In triangles ADC and CBA:  AB = DC; BAC = DCA and AC is common to both. So these triangles are congruent (SAS).

So AD = BC and AD // BC.

So the other pair of sides are equal and parallel.  That is the definition of a parallelogram.


#19 Re: Help Me ! » Help with proof or equation for daughter's algebra teacher » 2017-01-18 22:31:20

hi krissi772

Welcome to the forum.

Here's my suggestion.  I think it is Ok for a grade 8 student (I'm from the UK so I'm not so familiar with exactly what comes up when).

Imagine the seats have been numbered so that her ideal seat is x and his is y, where x+1 = y.

Then we need a standard algebra paradox.  You'll find it in many maths texts.

Let a be any number and b  so that a = b



Cancelling the (a-b) factors     ***   

Since b was  any number it could be:

So they are in adjacent seats!!!

*** I expect the teacher will want to quiz her some more about why this paradox is a 'cheat'.

The line that is illegal is the one marked ***

Cancelling a factor is the same as dividing by the factor.  (a-b) = 0 and it is not allowed to divide by zero. 

Here's why it is not allowed:

Anything times zero is always zero.  So:

If you divide by zero you get 63 = 89 which is clearly not a good basis for a number system.  To avoid this the definitions for the number system specify that you cannot divide by zero.

Hope that helps, smile


ps.  If the teacher still won't cooperate your daughter does have a plan B.  She can join MIF and get any help she wants here.

#20 Re: Help Me ! » Geometry: Consecutive sides that are congruent not a paralellogram? » 2017-01-18 21:31:05

Do you mean that one pair of adjacent sides are equal in length??  In that case there are lots of alternatives.  It could be a kite; or an arrowhead; or just an anyquad.  If you draw those two lines then you've fixed three out of four points.  Then put the fourth point anywhere you want.

If that isn't the problem, please explain what you mean here by congruent.  I would normally only use that term for a whole shape not just some lines.


#21 Re: Help Me ! » My son's problem » 2017-01-18 21:06:31

hi LC

The graph he needs is called a distance / time graph.

Mark hours on the across axis, 0 to 15 should do it.

On the up axis put distance from the dock, 0  to 250 should be enough.

Start the ship at (0,0) and also mark a point at (5,100) to show the ship's position when the latecomers arrive at the dock.

Join these points and continue it as the ship will carry on sailing away.

The latecomers are at (5,0) as they are at the dock, but 5 hours after the ship sailed.

If they use the speedboat they will travel 200 miles in 5 hours.  That will get them to the point (10, 200) so mark that point and join up their line of travel.  Where the two lines cross is the catch up moment and place.

If they wait 3.5 hours that will make the time 8.5 hours so mark a point at (8.5, 0).  One hour later the helicopter can be at (9.5,90) so mark that point, Draw the line of travel for the helicopter and extend it until it cuts the ship's line.  That will give the alternative catchup point.

So you can then decide which to go for.  Hidden is my attempt. The helicopter line is not quite accurately done but it'll give you an approximation.  Get him to try it first and then you can check it yourself. 

The graph should be enough to make the decision.  The catchup time for the speedboat option is easy to calculate.

For the helicopter it will require some algebra. Use the formula distance = speed x time   Call the meetup time T.  The ship has travelled 20T as it is going at 20 mph.  The helicopter starts at 8.5 and ends at T so its time of travel is (T-8.5).  It is going at 90 mph so it will travel 90.(T-8.5)

If you put these two distances equal (we want the helicopter to be at the same spot as the ship) and solve for T you will have the catchup time for the helicopter.

I would do both these calculations as it 'proves' by calculation which is the better option, and avoids any graphical inaccuracies.


ps.  I'm an ex-teacher, so naturally, I want the rest of his group to do the work too.  smile  If you think he has got it right, he could sit them down and instruct them in how to solve the problem.  This is a good way to test if you have understood something.  If you can teach others to do it, then you must have got the idea and understood it well.  If they won't cooperate he should submit his answer and tell the teacher the others haven't contributed.  Then it's up to the teacher to deal with them.  They may suffer a short term annoyance but in the end will respect him more (I hope; but if not, then they weren't very good friends any way)

#22 Re: Help Me ! » A peculiar enigma about a quadrilateral » 2017-01-18 20:36:13

hi Mathegocart

Draw any line EF and construct right angles at E and at F.

Choose any point G on the line from F and measure an angle of 67.

Extend this line to cut the other line at 113.  It will because the angle sum of a quadrilateral is 360.

This drawing is not accurately drawn.



#23 Re: Introductions » Hi » 2017-01-16 21:08:38

hi Maharadja

Welcome to the forum.


#24 Re: Help Me ! » The Perplexing Dimensions of The Rhombus. » 2017-01-16 21:02:49

hi Mathegocart

HE is a line of symmetry for the rectangle, the rhombus and the circle.  So proceed like this:

Find the radius OL = OF.

The diagonals of a rhombus bisect at right angles, so EOG = 90 and also OE // GF and  OG // EF.  (// = parallel)

=> Triangles GOF and FGE and congruent, so GE = OF = radius.


#25 Re: Help Me ! » cone » 2017-01-13 23:02:19

hi markosheehan

For part 2:

As you want dA/dt I would work with dV/dt = dV/dA  .  dA /dt

As D = 2 when H = 2. => r = 0.5h.

Express both V and A in terms of h and eliminate h to get V in terms of A.

Differentiate that.


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