The Delta-Wye (Δ-Y) Transformation
A star arm is 1/3 of a balanced triangle side
On the left is a balanced triangle (Δ) whose three sides are all R_Δ = 9Ω, and on the right a star (Y) whose three arms are all equal to R_Y. Both networks share the same three terminals A, B and C, and for them to be indistinguishable from outside, the resistance between any two terminals must match. The triangle is fixed at 6Ω between terminals (= 2/3·R_Δ), and the star gives 2·R_Y. Slide R_Y to find the value that makes the star look exactly like the triangle. When it matches, the resistors of both networks and the ≡ sign in the middle turn gold.
Circuits that series-parallel can't solve
In some layouts, such as bridge circuits, no resistor forms a pure series or parallel pair with any other. When a Wheatstone bridge is unbalanced, the middle branch blocks series-parallel reduction outright. If you then swap a triangle (Δ) group sharing three terminals for a star (Y) group, or the reverse, the blocked series-parallel path opens again. As long as they behave identically at the three outer terminals, the two forms are exactly the same part as far as the rest of the circuit can tell.
Match them at the terminals
Measure between two terminals of the balanced triangle and the side R_Δ joining them is in parallel with 2R_Δ, the other two sides in series. So R_Δ ∥ 2R_Δ = (R_Δ·2R_Δ)/(3R_Δ) = (2/3)R_Δ. In the balanced star, two terminals are joined just by two arms in series, giving 2R_Y. For the networks to look the same, 2R_Y = (2/3)R_Δ, that is, R_Y = R_Δ/3, which is why a 9Ω triangle and a 3Ω star cannot be told apart at the terminals. The star's arm is smaller because it covers only one leg from a terminal to the center, and it takes two such legs to span two terminals.
The general formula and going back
A triangle with unequal sides converts to a star by the same principle. One arm of the star is the product of the two triangle sides touching that terminal divided by the sum of all three sides: R_a = R_ab·R_ca/(R_ab+R_bc+R_ca). With equal sides the product is R_Δ² and the sum is 3R_Δ, returning R_Δ/3. Going back from star to triangle, multiply the arms in pairs, add the products, and divide by the arm opposite that side. Open the upper triangle of an unbalanced Wheatstone bridge into a star with this transform, and the rest of the circuit unravels cleanly by series-parallel.
Back to the first screen
As you slid the star arm R_Y, the star's terminal resistance 2R_Y grew, and at R_Y = 3Ω it matched the triangle's 6Ω exactly, turning the resistors of both networks and the ≡ sign gold. The 9Ω side of the balanced triangle folded into a star arm of 3Ω, 1/3 of it. The reason was simple: between terminals the triangle is one side parallel to two sides, (2/3)R_Δ, and the star is two arms in series, 2R_Y, so setting them equal gives R_Y = R_Δ/3. This one swap is the key that unlocks a bridge stuck under series-parallel.