Some MathZoom Induction Problems and 2011 USAJMO #5
by djmathman, Nov 23, 2012, 2:11 AM
Induction Problem 1 wrote:
Let 
, and let the sequence 
satisfy 
, 
for 
. Show that 
for all 
.
, and let the sequence 
satisfy 
, 
for 
. Show that 
for all 
.
for all 
since 
is positive. Additionally, note that 
.
,
produces 
.
,
,
.
,
.
Induction Problem #2 wrote:
Given 
, and 
for 
, show that 
is an integer for all .
, and 
for 
, show that 
is an integer for all .
for all 
and 
.
is an integer for all 
.
term. Suppose 
,
.![\[a_n+2a_{n-1}+\dfrac{a_{n-1}^2+1}{a_n}=a_n+2a_{n-1}+a_{n-2}=a_{n+1}+a_n.\]](http://latex.artofproblemsolving.com/f/9/e/f9eaf0a50de3567a8a73a28d6e2d8b385c5aedeb.png)
Similarly, when 
2011 USAJMO #5 wrote:
Points 
, 
, 
, 
, 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) , , are collinear, and (iii) 
. Prove that 
bisects 
.
, 
, 
, 
, 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) , , are collinear, and (iii) 
. Prove that 
bisects 
.![[asy]
size(250);
defaultpen(linewidth(0.8));
pair P=(3,0),C=dir(130),X=(1,0);
path circTan = circle((1.5,0),1.5),line=P--C;
real slope=(P.y-C.y)/(P.x-C.x);
pair x[]=intersectionpoints(unitcircle,circTan),A=intersectionpoint(line,unitcircle),pointbwDE=(x[1].x-2,x[1].y-2*slope);
path lineDE=pointbwDE--x[1];
pair E=intersectionpoint(lineDE,unitcircle),M=extension(A,C,x[0],E);
draw(unitcircle^^x[0]--P--x[1]--E--cycle^^C--P^^M--origin--A^^P--origin--x[0]);
dot(origin);
label("$A$",A,dir(origin--A));
label("$B$",x[0],dir(origin--x[0]));
label("$C$",C,dir(origin--C));
label("$D$",x[1],dir(origin--x[1]));
label("$E$",E,dir(origin--E));
label("$P$",P,dir(origin--P));
label("$M$",M,dir((C+x[0])/2));
label("$O$",origin,S);
label("$X$",X,NE);
[/asy]](http://latex.artofproblemsolving.com/b/7/5/b7556e025b1584d7baffa0e97d77d0a4be47a87d.png)
be the center of the circle, let 
,
be the point for which 
intersects circle 
and 
,
is a kite, which means that 
,
and 
have the same measure. Thus the measure of arc 
,
,
.
,
,
.
is cyclic, so 
.
,
is the midpoint of 
,
bisects This post has been edited 1 time. Last edited by djmathman, Apr 6, 2015, 3:14 AM
Reason: align tags
Reason: align tags
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