British FST-2 2014
by liberator, Aug 18, 2014, 6:26 PM
The British FST-2 is the second paper of the first round of the British TST, taken on the last day of a training camp held for the top 20 pupils nationally. In conjunction with FST-1 and several other areas of data, it is used to decide the squad of eight/nine for the final stage of IMO selection.
Problem 4: Let
be a triangle. For a point 
of the plane, let 
be the foot of the perpendicular dropped from to 
. Points 
are defined analogously. Find the locus of points in the plane such that ![\[PA \cdot PA' = PB \cdot PB' = PC \cdot PC'.\]](//latex.artofproblemsolving.com/8/7/7/877e1683af51a6504bca78b51c46247dd6670167.png)
Problem 6: Consider an odd prime
satisfying
![\[2^h \not \equiv 1 \pmod{p}, \forall 0 < h < p-1,\]](//latex.artofproblemsolving.com/0/f/1/0f177f0fe373ebdc96520a108ce45f20b22a7201.png)
and an even integer
with 
.
Consider the sequence
defined by:
![\[a_0 = a, \\ a_{n+1} = p - b_n, n \geq 0,\]](//latex.artofproblemsolving.com/4/f/2/4f263cac445128f669b5fa6cac1c190ebe51b381.png)
where
is the greatest odd divisor of 
. Show that 
is periodic, and find its smallest positive period.
Problem 4: Let
be a triangle. For a point 
of the plane, let 
be the foot of the perpendicular dropped from to 
. Points 
are defined analogously. Find the locus of points in the plane such that ![\[PA \cdot PA' = PB \cdot PB' = PC \cdot PC'.\]](http://latex.artofproblemsolving.com/8/7/7/877e1683af51a6504bca78b51c46247dd6670167.png)
![[asy]
/* British FST-2 2014 Problem 4, free script by liberator, 18 August 2014 */
unitsize(3cm);
defaultpen(fontsize(10pt));
/* Initialize objects */
pair A = dir(120);
pair B = dir(200);
pair C = dir(340);
pair P = (0.1, 0);
pair X = bisectorpoint(B,A,C);
pair Y = bisectorpoint(C,B,A);
pair Qa = reflect(A,X)*P;
pair Qb = reflect(B,Y)*P;
pair Qa1 = rotate(180, Qa)*A;
pair Qb1 = rotate(180, Qb)*B;
pair Q = intersectionpoint(A--Qa1, B--Qb1);
pair Ap = foot(P,B,C);
pair Bp = foot(P,C,A);
pair Cp = foot(P,A,B);
/* Draw objects */
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(P--Ap, rgb(0.4,0.6,0.8));
draw(P--Bp, rgb(0.4,0.6,0.8));
draw(P--Cp, rgb(0.4,0.6,0.8));
draw(P--A, rgb(0.4,0.6,0.8));
draw(P--B, rgb(0.4,0.6,0.8));
draw(P--C, rgb(0.4,0.6,0.8));
draw(Q--A, rgb(0.4,0.6,0.8));
draw(Q--B, rgb(0.4,0.6,0.8));
draw(unitcircle, rgb(0.9,0,0));
/* Draw anglemarks */
draw(anglemark(P,A,C), rgb(0.4,0.6,0.8));
draw(anglemark(B,A,Q), rgb(0.4,0.6,0.8));
draw(anglemark(C,B,P), rgb(0.4,0.6,0.8)); draw(anglemark(C,B,P,7), rgb(0.4,0.6,0.8));
draw(anglemark(Q,B,A), rgb(0.4,0.6,0.8)); draw(anglemark(Q,B,A,7), rgb(0.4,0.6,0.8));
draw(rightanglemark(A,Bp,P,2), rgb(0.4,0.6,0.8));
draw(rightanglemark(P,Ap,B,2), rgb(0.4,0.6,0.8));
draw(rightanglemark(P,Cp,A,2), rgb(0.4,0.6,0.8));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(A));
dot(B); label("$B$", B, dir(B));
dot(C); label("$C$", C, dir(C));
dot(P); label("$P$", P, 1.2*dir(30));
dot(Q); label("$Q$", Q, dir(Q));
dot(Ap); label("$A'$", Ap, dir(Ap));
dot(Bp); label("$B'$", Bp, dir(Bp));
dot(Cp); label("$C'$", Cp, dir(Cp));
[/asy]](http://latex.artofproblemsolving.com/9/1/d/91dab17bf070a2059ddf3ab8a71431b11abeb929.png)
be the isogonal conjugate of 
.
,![\[\frac{PA'}{PB'} = \frac{PB \cdot \sin \angle PBC}{PA \cdot \sin \angle PAC} = \frac{PB}{PA} \implies \sin \angle QBA = \sin \angle QAB.\]](http://latex.artofproblemsolving.com/4/e/0/4e0770e7913963fb9a1f319822cb981b2e58769a.png)
,
,
,Problem 6: Consider an odd prime
satisfying![\[2^h \not \equiv 1 \pmod{p}, \forall 0 < h < p-1,\]](http://latex.artofproblemsolving.com/0/f/1/0f177f0fe373ebdc96520a108ce45f20b22a7201.png)
and an even integer
with 
.Consider the sequence
defined by:![\[a_0 = a, \\ a_{n+1} = p - b_n, n \geq 0,\]](http://latex.artofproblemsolving.com/4/f/2/4f263cac445128f669b5fa6cac1c190ebe51b381.png)
where
is the greatest odd divisor of 
. Show that 
is periodic, and find its smallest positive period.
.
for non-zero 
.
,![\[\therefore \exists i,j \in \mathbb{N}: i > j, a_i = a_j \implies a_{i+1} = a_{j+1}, a_{i+2} = a_{j+2}, \cdots .\]](http://latex.artofproblemsolving.com/f/9/0/f90bff1ad3fa56c7e1d756105be66033e0aa7e9d.png)
and 
,
.
.![\[1 \leq \frac{a_{i-1}}{a_{j-1}} < 2,\]](http://latex.artofproblemsolving.com/1/8/0/180661062a90d8980881abdfa154ea48f16beeb2.png)
so 
.
.
is the minimal period. For convenience, let 
.![\[\prod_{i=0}^{2m-1} a_i = 2^{2r} \cdot \prod_{i=0}^{2m-1} b_i.\]](http://latex.artofproblemsolving.com/3/d/5/3d5909481b13267dff95c8c0f68177a2ef8cc023.png)
,
.![\[\prod_{i=0}^{2m-1} a_i \equiv (-1)^{2m} \cdot \prod_{i=0}^{2m-1} b_i \pmod p \implies 2^{2r} \equiv 1 \pmod p.\]](http://latex.artofproblemsolving.com/9/f/3/9f361807f0e6328cc8bf01c4b34f75ba0ddbaceb.png)
.![\[v_2 \left( \prod_{i=\frac{p}{2}}^{p} i \right) = \frac{p-1}{2} \implies r = v_2 \left( \prod_{i=0} ^ {m-1} a_i \right) \leq \frac{p-1}{2} \implies r= \frac{p-1}{2}.\]](http://latex.artofproblemsolving.com/7/4/4/7442aed4c226ac36572c2675e24c09a4e10d1b54.png)
comprises of all the even integers in the interval ![$[\tfrac{p}{2},p]$](http://latex.artofproblemsolving.com/c/c/b/ccbe9de577be90a34aa2152aee7426f5c938c50e.png)
.![\[m = \left \lceil \frac{p-1}{4} \right \rceil .\]](http://latex.artofproblemsolving.com/f/5/c/f5ca51a93e87a57366c056ad73545e260d54e479.png)
This post has been edited 3 times. Last edited by liberator, Sep 8, 2014, 7:37 PM
October 2018
April 2018