Problem:
Quadrilateral APBQ is inscribed in circle ω with ∠P=∠Q=90∘ and AP=AQ<BP. Let X be a variable point on segment PQ. Line AX meets ω again at S (other than A ). Point T lies on arc AQB of ω such that XT is perpendicular to AX. Let M denote the midpoint of chord ST. As X varies on segment PQ, show that M moves along a circle.
Solution:
Let O denote the center of ω, and let W denote the midpoint of segment AO. Denote by Ω the circle centered at W with radius WP. We will show that WM=WP, which will imply that M always lies on Ω and so solve the problem.
We present two solutions. The first solution is more computational (in particular, with extensive applications of the formula for a median of a triangle); the second is more synthetic.
Set r to be the radius of circle ω. Applying the median formula in triangles APO,SWT,ASO,ATO gives
4WP24WM22WS22WT2=2AP2+2OP2−AO2=2AP2+r2=2WS2+2WT2−ST2=AS2+OS2−AO2/2=AS2+r2/2=AT2+OT2−AO2/2=AT2+r2/2
Adding the last three equations yields 4WM2=AS2+AT2−ST2+r2. It suffices to show that
4WP2=4WM2 or AS2+AT2−ST2=2AP2(1)
Because XT⊥AS,
AT2−ST2=(AX2+XT2)−(SX2+XT2)
=AX2−SX2=(AX+XS)(AX−XS)=AS(AX−XS)
It follows that AS2+AT2−ST2=AS2+AS⋅(AX−XS)=AS2+AS(2AX−AS)= 2AS⋅AX, and (1) reduces to AP2=AS⋅AX, which is true because triangle APX is similar to triangle ASP (as ∠PAX=∠SAP and ∠APX=arc(AQ)/2=arc(AP)/2=∠ASP).
OR
In the following solution, we use directed distances and directed angles in order to avoid issues with configuration (segments ST and PQ may intersect, or may not as depicted in the figure.)
Let R be the foot of the perpendicular from A to line ST. Note that OM⊥ST, and so ARMO is a right trapezoid. Let U be the midpoint of segment RM. Then WU is the midline of the trapezoid. In particular, WU⊥RM. Hence line WU is the perpendicular bisector of segment RM. It is also clear that AW is the perpendicular bisector of segment PQ. Therefore, W is the intersection of the perpendicular bisectors of segments RM and PQ. It suffices to show that quadrilateral PQMR is cyclic, since then W must be its circumcenter, and so WP=WM.
(To be precise, this argument fails when ST and PQ are parallel, because then R=M and the perpendicular bisector of RM is not defined. However, it is easy to see that this can happen for only one position of X. Because the argument works for all other X, continuity then implies that M lies on Ω for this exceptional case as well.)
Let lines PQ and ST meet in V. By the converse of the power-of-a-point theorem, it suffices to show that VP⋅VQ=VR⋅VM. On the other hand, because PQTS is cyclic, by the power-of-a-point theorem, we have VP⋅VQ=VS⋅VT. Therefore, we only need to show that
VS⋅VT=VR⋅VM(2)
Note that M is the midpoint of segment ST. Then (2) is equivalent to
2VS⋅VT=VR⋅(2VM)=VR⋅(VS+VT)
or
VS⋅VT−VS⋅VR=VT⋅VR−VT⋅VS
or equivalently
VS⋅RT=VT⋅SR or SRVS=RTVT(3)
We claim that XS bisects ∠VXR. Indeed, because AB is the symmetry line of the kite APBQ,AB⊥PQ, and so ∠VXS=∠QXA=90∘−∠XAO=90∘−∠SAO. Because O is the circumcenter of triangle AST,
∠VXS=90∘−∠SAO=∠ATS
On the other hand, because ∠AXT and ∠ART are both right angles, quadrilateral AXRT is cyclic, implying that ∠SXR=∠ATR=∠ATS. Our claim follows from the last two equations.
Combining our claim and the fact that XS⊥XT, we know that XS and XT are the interior and exterior bisectors of ∠VXR, from which (3) follows, by the angle-bisector theorem. We saw that (3) was equivalent to (2) and that this was enough to show that PQMR is cyclic, which completes the solution, so we are done.
The problems on this page are the property of the MAA's American Mathematics Competitions
