Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 19H / a

Table of contents
- Solution a

Solution

Let

D, E, F

be the midpoints of

A B, A C, BC

. Write

a = E_{A} T_{BC}, d = E_{D} T_{BC} = E_{E} T_{BC}, f = E_{F} T_{BC} .

First-step analysis for the simple random walk gives

a = 1 + d,

because

A

has neighbours

D, E

;

d = 1 + \frac{a + d + f}{4},

because

D

has neighbours

A, B, E, F

; and

f = 1 + \frac{d}{2},

because

F

has neighbours

B, C, D, E

. Solving,

d = 4, f = 3, E_{A} T_{BC} = a = 5 .

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