Melody

Can someone answer this please?

Suppose that the graph of a certain function, y=f(x), has the property that if it is shifted 20 units to the right, then the resulting graph is identical to the original graph of y=f(x).  What is the smallest positive a such that if the graph of y=f(x/5)$ is shifted a units to the right, then we know that the resulting graph is identical to the original graph of $y=f(x/5)?

The answer is 5*20=100

This is when we know for sure that the  translation of a (in the right direction) will make the graph identical.

Here is a simple example

In a right triangle with integer length sides, the hypotenuse has length 39 units. How many units is the length of the shorter leg?

\(x^2+y^2=39^2\\x^2+y^2=1521 \)

sqrt(1521-38^2) = 8.7749643873921221

sqrt(1521-37^2) = 12.3288280059379529

sqrt(1521-36^2) = 15

there you go your short sides are  36 and 15

there is probably only one answer but i have not checked that.

The shortest side is 15units

You have the setting out of these off pat Chis :)  I am impressed!

As you know I have given up and just do them on paper and take a photo. 

It says to leave it as a factored expresssion so I think you only need to say

(x-53)(x-22)=0

and leave it as that.

Thanks Heureka, a great one for me to study :)

Much appreciated :)

A factor of what?  You should read your question after you post it then you would know it is not complete!

Thanks X^2 

What you say makes sense but there is still room for confusion ://

Thanks for complementing me on my answer Chris. 

I'm sure you know better than most that when we put a lot of effort into an answer like this we do want someone to notice.

I mean we get our own satisfaction but still it is nice if we have a small appreciative audience as well.   

You provide so many excellent geometry answers, I did not think you would notice this one.  

".I did not know the thing about the exterior angle of a cyclic quad = the opposite interior angle"

I kind of extrapolated that ...

One of the most important features of a cyclic quad is that opposite angles are supplementary.

I was going to use that.

But then I realised that since this is true it means, by extension, that the exterior angle of a cyclic quad is equal to the opposite internal angle. 

It just meant that I could skip one step in the proof, that is all. :)

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With these proofs I am sometimes a bit confused about when I should use the 'congruent to' sign and when I should just use the equal sign. Do you have any confusion over this?

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Did you see Rosala's question about series that are combinations of Arithmetic Progressions and Geometric progressions?

She asked for the formula derivation to be explained and then she had 2 questions using it.

I answered the first, but I couldn't do the second.

It was 'new' maths for me and quite interesting. 

I'll see if I can find the link.

Arr, I can see Heureka is answering it now :))

I have labeled the points T, X and R as shown in my diagram.

I have also let  \(\angle DPX =\alpha\)   \(\;\;and\;\;\angle AQX=\beta\;\;and \;\;\angle ADC=\theta \) 

Now:

\(\angle DPX\cong \angle CPX= \alpha \qquad \qquad \overline{PX}\;\;bisects \angle DPC\\ \angle AQX\cong \angle DQX= \beta \qquad \qquad \overline{QX}\;\;bisects \angle AQD\\~\\ \angle ADC\cong \angle QBR=\theta \\ \qquad \text{Exterior angle of cyclic quad= opposite internal angle}\\~\\ \angle BRX=\angle QBR+\angle BQR=\theta + \beta\\ \qquad \text{Exterior angle of triangle = sum of opp internal angles in } \triangle BRQ\\~\\ \angle ATX=\angle TDQ+\angle TQD = \theta+\beta\\ \qquad \text{Exterior angle of triangle = sum of opp internal angles in } \triangle TQD\\~\\ \therefore \angle BRX=\angle ATX\\\)

\(Consider\;\; \triangle PXT \;\;and\;\; \triangle PXR\\ \angle PTX=\angle PRX=\theta + \beta\\ \angle TPX=\angle RTX=\alpha PX = PX \;\; common side\\ \therefore \triangle PXT=\triangle PXR \\ \qquad \text{Two angles and corresponding side test.}\\~\\ \therefore PXT=\angle PXR\\ \qquad \text{Corresponding angles in congruent triangles}\\ But \;\;\angle PXT+\angle PXR=180^\circ \\\qquad \text{Adjacent supplementary angles}\\ \therefore \angle PXT=\angle PXR=90^\circ\\ \therefore QT\;\;and \;\;PX \text{ are perpendicular.}\)

\(\text{Therefore the angle bisectors of  DPC and  AQD are perpendicular.         QED}\)

[ Well that is assuming I have not put letters in stupid places anyway ]