Problem:
Let △ABC be a non-equilateral, acute triangle with ∠A=60∘, and let O and H denote the circumcenter and orthocenter of △ABC, respectively.
(1) Prove that line OH intersects both segments AB and AC.
(2) Line OH intersects segments AB and AC at P and Q, respectively. Denote by s and t the respective areas of triangle APQ and quadrilateral BPQC. Determine the range of possible values for s/t.
Solution:
(a): Without loss of generality, we assume that AB>AC. Set β=∠ABC and γ=∠ACB. We have β<60∘<γ and β+γ=120∘.
Note that ∠BAO=90∘−∠ACB=90∘−γ<90∘−β=90∘−∠ABC=∠BAH, and so AO lies inside ∠BAH. Similarly, ∠ABO=90∘−γ<30∘=∠ABH, and so BO lies inside ∠ABH. Hence O lies inside △ABH, and line OH intersects side AB. In the same way, ∠CAH=90∘−γ<90∘−β=∠CAO and ∠ACH=30∘<90∘−β=∠ACO; hence H lies inside △ACO, and line OH intersects side AC.
(b): The range of s/t is the open interval (4/5,1).
Based on (a), we may consider the configuration shown above. Note that ∠BOC=2∠BAC=120∘ and ∠BHC=180∘−∠HBC−∠HCB=180∘−(90∘−γ)−(90∘−β)=120∘, from which it follows that BOHC is cyclic. In particular, ∠POB=180∘−∠HOB=∠HCB=90∘−β, and it follows that
∠APQ=∠ABO+∠POB=(90∘−γ)+(90∘−β)=60∘
Since ∠PAQ=60∘ as well, we see that △APQ is equilateral.
Next note that ∠POB=90∘−β=∠ACO=∠QCO and ∠PBO=90∘−γ=∠HBC=∠HOC=∠QOC; since BO=OC, we have congruent triangles △BPO≅△OQC. Thus
AB+AC=AP+PB+CQ+QA=AP+QO+OP+QA=AP+PQ+QA
and so AP=PQ=QA=3b+c, where we write b=AC and c=AB. Therefore we have
By our assumptions that b<c and △ABC is acute, it follows that the range of m is 1<m<2. (One can see this, for instance, by having A move along the major arc BC from one extreme, where ABC is equilateral and c/b=1, to the other, where ∠ACB=90∘ and c/b=2, and noting that c increases and b decreases during this motion.) For m∈(1,2), the function f(m)=m+1/m is continuous and increasing: if 1<m<m′<2, then f(m′)−f(m)=mm′(m′−m)(mm′−1)>0. Thus the range of f(m) for m∈(1,2) is (f(1),f(2))=(2,25). It follows that the range of s+ts=92+f(m) is (94,21), and the range of ts is (54,1).
This problem and the first solution was suggested by Zuming Feng.
OR
(b): We use complex numbers. Let O=0,B=1,C=ω=e2πi/3, and A=a with ∣a∣=1. Then H=1+ω+a=a−ω2. Bearing in mind that the equation for the line through complex numbers w1 and w2 is w2−w1z−w1=w2−w1zˉ−w1 (i.e., the quotient w2−w1z−w1 is purely real), we see that P, which is the intersection of AB and OH, lies at the point z satisfying
a−1z−1=aˉ−1zˉ−1 and a−ω2z=aˉ−ωzˉ
Substituting aˉ=1/a, eliminating zˉ, and solving for z yields z=1−ωa+1. Thus the vector AP is given by the complex number 1−ωa+1−a=1−ωaω+1. Similarly Q lies at the point ω−1aω+ω2 and the vector AQ is ω−1a+ω2. It follows that AP=31∣ωa+1∣=31∣∣∣a+ω2∣∣∣=AQ.
Now AB=1−a is collinear with AP=1−ωaω+1, and the ratio of the lengths of these vectors is APAB=(1−a)/(1−ωaω+1)=aω+1(1−a)(1−ω); similarly AC=ω−a is collinear with AQ=ω−1a+ω2, and AQAC=a+ω2(ω−a)(ω−1)=aω+1(ω−a)(ω2−ω). Thus
