The terminology can be confusing for a newbie, actually. The term "virtual short circuit" refers to the fact that in an opamp circuit with negative feedback the circuit is arranged in a way that (ideally) makes the voltage across the two opamp inputs zero.
Since one of the properties of a short circuit between two points is that the voltage across those points is zero, the people who invented that terminology considered (I guess) an intuitive thing to call what happens between the input terminals of the opamp a "virtual short". They called it "virtual" because it lacks the other property of a real (ideal) short: to gobble up any amount of current without problems! Alas, that's no small difference! They could have called the thing in a less confusing manner ("the principle of voltage balancing"!?!), but "the virtual short principle" sounds cooler, probably! Who knows?!
So, when we say that between the two inputs there is a virtual short, it's just an easy and conventional way to say that the circuit strives to balance the voltages at the inputs, i.e. it tries to make them and keep them equal.
Note that the existence of the "virtual short" is a property of the circuit, not of the opamp (although it exploits the ideally infinite gain of the opamp), whereas the fact that no current flows into the inputs is a property of the opamp (ideally).
EDIT (prompted by a comment)
I'll try to be clearer about what I said above. The virtual short is exclusively due to two key factors combined together: very high gain + negative feedback.
Let's do some math do convince ourselves. Let's call \$V^+\$ and \$V^-\$ the voltages at the non-inverting and the inverting inputs of the opamp, respectively, and \$V_o\$ the output voltage. A real opamp, in this respect, is a differential amplifier, i.e. \$V_o = A(V^+-V^-)\$, where \$A\$ is the open-loop gain of the opamp.
Inverting that relationship you get \$V^+-V^-=V_o/A\$. Thus, for finite \$V_o\$ and infinite \$A\$, you get that the difference between the inputs becomes zero.
Where did the negative feedback play a role? Nowhere, till now!!! The catch is that a real opamp needs negative feebdback to keep its output from saturating, in which case the simple linear model of the opamp (i.e. that gain formula) would no longer apply, except outside a very small interval of input voltages (assuming a classic non-inverting configuration where \$V^+\$ is the input voltage and \$V^-\$ is a fraction of the output).
Apply negative feedback and you'll get a zero differential voltage at the inputs over a meaningful range of input voltages.
pjc50 summed up the solution in comments.
Input impedance is defined as the impedance you would see no matter what the input voltage/current. With the fact that ideal op-amp analysis pins both inputs of the op-amp to the same voltage, and one of the inputs pinned to ground, the other terminal can be treated as a virtual ground. Now you have an infinite current source/sink at that virtual ground. That means if you create a test input voltage on the input side of the circuit, all it will see is 10k to "ground". 1V/10k=100uAmps. Input impedance is the test voltage divided by the resulting current so 1V/100uAmps = 10k. So that's how they're coming up with 10k as an input impedance.
If you wanted to determine the output impedance, you could look at the tail end and try and inject a 1V supply at the output side of the op-amp. The output of an op-amp can source or sink infinite current (pseudo-ground) so you basically have a connection from the output wire directly to ground. A wire directly to ground is 0 ohms so the output impedance in this case is 0 (or very low).
From what I can tell, your confusion comes then from not understanding how an ideal op-amp works. The input side of an op-amp tries to pin its two inputs to the same voltage and the output can source/sink infinite current to create the desired voltage.
Best Answer
Yes, this is the circuit input impedance (between the left Rin end and ground) that is different from the op-amp input impedance (between its two inputs).
Really, the op-amp input impedance is infinite... but if the op-amp was standalone. Note that here a network consisting of two elements in series (Rf and the op-amp output) shunts the op-amp differential input (ie, it is connected between them). So this network determines the resistance between the op-amp inputs. Let's see what its resistance is...
The input current flows through Rf and creates a voltage drop VRf = Iin.Rf. To keep the voltage at the inverting input equal to zero (obbeying the so-called "golden rule"), the op-amp adjusts its output voltage equal to the voltage drop across Rf (Vout = -Iin.Rf) and adds it in series. The op-amp output serves as a variable voltage source that copies the voltage drop across Rf and removes it. As a result, the differential voltage between the op-amp inputs is zero (Iin.Rf - Iin.Rf = 0). Thus there is current flowing but there is no voltage... so the resistance of this network (and between the op-amp inputs) is zero... virtual zero. Figuratively speaking, the inputs are short connected by something like a "piece of wire".
So, the conclusion is that the circuit input impedance is determined only by Rin.
The conceptual picture below illustrates my explanations. Pay attention to something very important for understanding the circuit - the four elements (two voltage sources and two resistors) are connected in a loop and the same current flows through them (its trajectory is drawn in green). Also note another very important property of this configuration - the two voltages Vin and Vout have the same polarity when travelling the loop; so they are summed according to KVL.
Try to grasp the idea; if you have any questions, I will be happy to answer. I know it will be a little difficult for you to understand my slightly unconventional explanations... but if you succeed, the benefits will be great for you... You will know what the secret of op-amp inverting circuits is. For example, you can easily answer a similar question.
In the edit below, I have exposed some basics of my philosophy about negative feedback circuits as a response to AnalogKid's updates.
UPDATE 1
Undisturbed follower. Although it is possible for an op-amp to change the voltages of both its inputs (for example, in an NIC), in most cases it only changes the voltage of its inverting input so that it (always) follows the voltage of its non-inverting input. The latter is permanently zero (in the case of the inverting amplifier) or is initially zero (in the case of the non-inverting amplifier). So, by its nature, the op-amp circuit with negative feedback is a zero voltage follower. Its simplest implementation consists of only one op-amp whose output is connected to its inverting input.
Disturbed follower. From now on, each new element inserted (resistor, capacitor, diode, transistor, etc.) or voltage or current applied (Vin, Iin) acts as a "disturbance" for this initial follower since it tries to change the zero voltage reference. The op-amp reacts to the disturbance to overcome it and we take its reaction as an output. In this way, all possible op amp circuits with negative feedback can be obtained by intentionally disturbing them.
I would say: Since the input voltage source pushes a current through R1 into the node but the op-amp draws the same current through R2 from the node (or the input source draws a current via R1 from the node but the op-amp pushes the same current via R2 to the node), the voltage of the node does not change.
From another point of view, this 4-element configuration can be seen as a balanced bridge.
UPDATE 2
Exactly! Strange as it may seem, it is this delay that makes it possible to explain the circuit operation. If we consider the op-amp as a device without delay (Vout = k.Vin), we fall into a vicious circle.
So, when Vin changes (eg, decreases), in the first moment, Vout does not change… and the voltage divider R1-R2 is driven from the left side by Vin. After a while, the op-amp responds to the change by starting to increase its output voltage… and now the voltage divider R1-R2 is driven from the right side by Vout.
UPDATE 3
Exactly! "Virtual ground" is a voltage source whose voltage is a "copy" of another (reference) voltage (that can be zero). Figuratively speaking, virtual ground is a clone of another but real ground.
In 2007, I put a lot of effort into finding out what virtual ground really is and telling it on Wikipedia. Here is an old revision of the page and a heated discussion on the talk page. In the end, the page was trimmed and is now in a miserable state (Wikipedia EE is another place where there are terrible people; the only strange thing is how they are allowed to run wild).
COMMENT 1
Original representation of the negative feedback arrangement that I have never seen! I use two approaches to presenting it:
Amplifier. If the amplifier used to build the negative feedback circuit has a relatively small gain, I consider it as an amplifier with some moderate gain; then its input voltage cannot be ignored. This is the case in transistor circuits.
Integrator. If the amplifier used to build the negative feedback circuit has an extremely high gain, I simply consider it as an integrator… and I do not talk about gain at all.This is the case in op-amp circuits.