You may find it helpful to draw your own diagram as you work through this solution.

Consider triangle AXB. If this has a right angle at X then by Pythagoras' Theorem,

AX² + BX² = AB²

or, AX² + BX² = 100.

Now consider the triangles AMX and BMX.

Let angle AMX be a, so angle BMX is 180 - a.

In triangle AMX, using the cosine rule, we can use the cosine rule to find that

AX² = 5² + MX² - 2 x 5 x MX x cos a = 25 + MX² - 10 MX cos a (*)

Similarly, in triangle BMX, applying the cosine rule gives

BX² = 5² + MX² - 2 x 5 x MX x cos(180 - a) = 25 + MX² - 10 MX cos (180 - a)

Now recall* that cos (180 - a) = - cos a,

* or take a look at the symmetry of a cosine graph!

so BX² = 25 + MX² + 10 MX cos a (**)

Adding (*) and (**), we get

AX² + BX² = 50 + 2MX².

But we know that AX² + BX² = 100, so it must be that

50 + 2MX² = 100

2MX² = 50

and so MX = 5.

Now we just need to find the length of MC, which is the hypotenuse of teh right-angled triangle AMC.

MC² = 10² + 5² = 125, therefore MC = 5√5

As XC = MC - MX, XC = 5√5 - 5.

Therefore MX : XC = 5 : 5√5 - 5 = 1 : √5 - 1.

Answer to bonus question

This can be found using exactly the same method that is shown above! The only difference is that now XC = MC + MX = 5√5 + 5.

Therefore MX : XC = 5 : 5√5 + 5 = 1 : √5 + 1.

Challenge 361: Right-angle Ratio