Problem 3

Let

ABC

be an equilateral triangle. Let

P

be a point on

AC

and

Q

a point on

AB

so that triangles

ABP

and

ACQ

are acute. Let

R

and

S

be the orthocentres of triangles

ABP

and

ACQ

, respectively. Let

T

be the common point of the segments

BP

and

CQ

. Find all possible values of

\angle CBP

and

\angle BCQ

such that triangle

TRS

is equilateral.

Step 2 of 8: Prove the comparison lemma

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\alpha>\beta\implies RT>ST.

Detailed analysis

Let the three altitudes of the equilateral triangle meet at its centre G. The points P and Q lie on the corresponding altitude segments, so T lies in the relevant central triangle. Project R and S onto the altitude through A and onto BC as in the official solution. Equal altitude lengths give GS<GR when alpha>beta; the two projection inequalities then imply RT>ST.