This isn't for the Arduino, it's for the Lego NXT

See only now do you tell us the vital bit of information, so only now are we in a position to supply a sensible answer!

You need to apply kershoffs law, the sum of all currents into a node is equal to the sum of currents out of a node.

So if your potential divider has 10K resistors, with a voltage n, then into the node at the analogue input you have a current of

n / (10K + 10K) = 0.05n mA

Add a resistor of 10K to a 4.7V reference voltage, then the current through that will be defined by the voltage across it. This is 4.7V - n/2 so will contribute:-

(4.7 - n/2) / 10K

This will then add to the current flowing down R3 and increase the voltage across it, raising the current at the node and thus lowering the current contribution of R1.

So you can work this out by iteration, or by calculus, or in fact you could just measure it.