Input Impedance of the Inverting Op Amp
In a traditional inverting opamp circuit, where ((all)) of the operational parameters are within their ranges, the inverting input node behaves as if it were galvanically connected directly to GND.  It isn't, of course, but it can mimic a true ground connection so well that it has a name - Virtual Ground - and its own Wiki page:
https://en.wikipedia.org/wiki/Virtual_ground
UPDATE 1:
The mimicking happens because, in a closed-loop, linear system, the opamp does whatever it takes to keep its two inputs at the same voltage.  So whatever the current is through Rin, the opamp drives Rf such that an equal but opposite (because it is the ((inverting)) input) current flows into the node, and the voltage at the Rin-Rf node just sits there at 0 V (or whatever voltage the non-inverting input is tied to).
UPDATE 2:
Because there is a real, physical delay between the inputs and output of an opamp, closing a negative feedback loop works like this.  The input changes; let's say it goes a bit negative.  Now, the inverting input is less than the non-inverting input; or, the + input is more positive.  When the + input goes up, the output goes up (the definition of &quot;non-inverting&quot;).  This increases the voltage at the right side of Rf, which increases the voltage at the left side of Rf which is the inverting input, making it equal to the non-inverting input again. Since the non-inv input is connected to GND, the effect is that even though the input signal changed, the voltage at the inv input did not; it appears stuck to GND.
Here is where the controllable gain happens.  Because Rf and Rin form a two-resistor voltage divider, they control how much the output has to change to pull the input back to a zero voltage difference.  Example: Rin = 10K; Rf = 100K.  In order to hold the center node (which is connected to the - input) at 0 V (OR (important) at any other fixed voltage), if the left side of the divider goes down 1 V, then the right side has to go up xxx V to maintain the center node at no change.  If the input signal goes down to -1 V, the opamp output goes up.  If it goes to only +9 V, the - input still is lower than the + input by 0.9091 V so the output keeps going more positive.
Remember that there is a delay time between when the input changes and when the output &quot;catches up&quot;.  When the output passes 10 V and everything should be balanced, it takes a while for that input condition to make it though the opamp circuitry and affect the output, and diring that time the output keeps increasing.  If the output goes to 11 V, now the - input is more positive then the + input by 0.9091 V, and this drives the opamp output to go more negative.  And up and down and up and down.  This is a decreasing-amplitude sinusoidal oscillation called ringing; it eventually decays to 0 V, and is very visible when an opamp is hit with a very fast input signal change.  &quot;Normal&quot; opamps have an internal lowpass filter (called &quot;compensation&quot;) so the output overshoots way less than 1 V, but many datasheets have a oscillograph showing the overshoot and/or damped ringing.
This entire analysis also can be worked through using the currents in the two resistors rather than the voltages across them.
UPDATE 3:
The reason it is called a virtual &quot;ground&quot; is that in traditional amplifier circuits with bipolar supplies, the non-inverting input is referenced to GND, usually through a resistor.  But a more pedantically correct term would be &quot;virtual reference potential&quot;.  Negative feedback drives the - input to equal the + input - whatever that voltage is.  This is how a two-resistor voltage divider at the + input holds the opamp output at Vcc/2 in a single-supply circuit.  But even with the DC offset, the invarience characteristic still is there.  &quot;&quot;&quot;The ground is strong in this one.&quot;&quot;&quot;