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The {(hhH)²} face has corner angles of h = 108° and H = 144°. Note that h is the same corner angle as the pentagon, thence that non-regular hexagon reuses 4 out of 5 vertices of such a virtually inscribed pentagon.

Incidence matrix according to Dynkin symbol

ofxfx3oofxx&#xt →   height(1,2) = sqrt[(3-sqrt(5))/6] = 0.356822
                    height(2,3) = 1/sqrt(3) = 0.577350
                    height(3,4) = height(4,5) = sqrt[(3+sqrt(5))/6] = 0.934172

o....3o....     | 1 * * * * | 3 0 0 0 0 0 0 0 | 3 0 0 0 0
.o...3.o...     | * 3 * * * | 1 2 0 0 0 0 0 0 | 2 1 0 0 0
..o..3..o..     | * * 6 * * | 0 1 1 1 0 0 0 0 | 1 1 1 0 0
...o.3...o.     | * * * 6 * | 0 0 0 1 1 1 0 0 | 0 1 1 1 0
....o3....o     | * * * * 6 | 0 0 0 0 0 1 1 1 | 0 0 1 1 1
----------------+-----------+-----------------+----------
oo...3oo...&#x  | 1 1 0 0 0 | 3 * * * * * * * | 2 0 0 0 0
.oo..3.oo..&#x  | 0 1 1 0 0 | * 6 * * * * * * | 1 1 0 0 0
..x.. .....     | 0 0 2 0 0 | * * 3 * * * * * | 1 0 1 0 0
..oo.3..oo.&#x  | 0 0 1 1 0 | * * * 6 * * * * | 0 1 1 0 0
..... ...x.     | 0 0 0 2 0 | * * * * 3 * * * | 0 1 0 1 0
...oo3...oo&#x  | 0 0 0 1 1 | * * * * * 6 * * | 0 0 1 1 0
....x .....     | 0 0 0 0 2 | * * * * * * 3 * | 0 0 1 0 1
..... ....x     | 0 0 0 0 2 | * * * * * * * 3 | 0 0 0 1 1
----------------+-----------+-----------------+----------
ofx.. .....&#xt | 1 2 2 0 0 | 2 2 1 0 0 0 0 0 | 3 * * * * {5}
..... .ofx.&#xt | 0 1 2 2 0 | 0 2 0 2 1 0 0 0 | * 3 * * * {5}
..xfx .....&#xt | 0 0 2 2 2 | 0 0 1 2 0 2 1 0 | * * 3 * * {(hhH)2}
..... ...xx&#x  | 0 0 0 2 2 | 0 0 0 0 1 2 0 1 | * * * 3 * {4}
....x3....x     | 0 0 0 0 6 | 0 0 0 0 0 0 3 3 | * * * * 1 {6}