A Few Personal Favorites from CMIMC (Part II)
by djmathman, Jan 30, 2017, 2:51 AM
Once again, here are a few of my favorite problems from this year's contest that I wrote along with their solutions. I think I wrote 29 questions out of the 75 that appeared on the test, which means I've broken 60 total problems over the two years; I'm not sure if this is sustainable....
Oops this list doesn't have much variety....
2017 CMIMC Team #2 wrote:
Suppose 
, 
, and 
are nonzero complex numbers such that 
. Compute ![\[(x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right).\]](//latex.artofproblemsolving.com/b/4/e/b4e212f566995443c545f40f7fc67ddeae14e83e.png)
, 
, and 
are nonzero complex numbers such that 
. Compute ![\[(x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right).\]](http://latex.artofproblemsolving.com/b/4/e/b4e212f566995443c545f40f7fc67ddeae14e83e.png)
,
,
.
and ![\[x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) + 3xyz = S_1^3-3S_1S_2 + 3S_3.\]](http://latex.artofproblemsolving.com/b/2/c/b2c3c4ac288e1df9aa8743086efe4b1edee8cfd1.png)
Thus the condition becomes ![\[S_1(S_1^2-2S_2) = S_1^3 - 3S_1S_2 + 3S_3\quad\implies\quad S_1S_2 = 3S_3.\]](http://latex.artofproblemsolving.com/a/5/e/a5e32710c5ffb7d22938aec3ecc22b37683b03da.png)
Hence 
,
gives the desired answer of 
.
and solving for 2017 CMIMC Geometry #3 wrote:
In acute triangle 
, points 
and 
are the feet of the angle bisector and altitude from 
respectively. Suppose that 
and 
. Compute 
.
, points 
and 
are the feet of the angle bisector and altitude from 
respectively. Suppose that 
and 
. Compute 
.
and 
.![\[\dfrac{DC}{AC} = \dfrac{DB}{AB} = \dfrac{DC-DB}{AC-AB} = \dfrac23.\]](http://latex.artofproblemsolving.com/f/8/7/f8789745e9ecd789522c1840db2df4fff791fdd0.png)
Thus 
and 
.
.![\[(EC-EB)(EC+EB) = (AC-AB)(AC+AB)\quad\implies\quad (EC-EB)\cdot\dfrac23(x+y) = 36(x+y).\]](http://latex.artofproblemsolving.com/e/d/a/eda72779f000e360453681628a443758df673854.png)
Simplification yields 
.2017 CMIMC Algebra #7 wrote:
Let 
, 
, and 
be complex numbers satisfying the system of equations 
Find 
.
, 
, and 
be complex numbers satisfying the system of equations 
Find 
.![\[E_r = \dfrac{a^r}{b+c}+\dfrac{b^r}{c+a}+\dfrac{c^r}{a+b}\]](http://latex.artofproblemsolving.com/2/e/4/2e4b30f560e3cb38e37db1dc3c870fc6ddd11ca6.png)
for all nonnegative integers 
.
This is this identity that will be the workhorse for our solution.
gives 
,
.
case gives 
.
case yields 
,
.
Finally, recall that 
implies 
,![\[11(5) - 16 = 39 = 3abc\quad\implies\quad abc=\boxed{13}.\]](http://latex.artofproblemsolving.com/a/b/9/ab9189e81812eed802a73679ae35c1e26e2876ee.png)
2017 CMIMC Geometry #9 wrote:
Let 
be an acute triangle with circumcenter 
, and let 
denote the point on 
for which 
. The circumcircle of 
intersects lines 
and 
for the second time at and respectively. Suppose that 
, 
, and 
are concurrent. If 
and 
, compute .
be an acute triangle with circumcenter 
, and let 
denote the point on 
for which 
. The circumcircle of 
intersects lines 
and 
for the second time at and respectively. Suppose that 
, 
, and 
are concurrent. If 
and 
, compute .
.![[asy]
import olympiad;
size(250);
defaultpen(linewidth(0.8));
pair A = dir(119.5), B = dir(215), C = dir(325), Q = (A.x,-1*A.y), P = intersectionpoint(A--Q,B--C), X = 2*B - A, O = origin, Y = circumcenter(B,O,C);
path cir = circumcircle(B,O,C);
pair[] Dp = intersectionpoints(A--C, cir), Ep = intersectionpoints(A--X,cir);
pair D = Dp[0], E = Ep[1];
draw(B--C--A--E--D^^circumcircle(A,B,C)^^A--Q);
draw(arc(Y,circumradius(B,O,C),-5,185));
draw(rightanglemark(A,P,B,1.5)^^E--O--D);
dot(O);
draw(circumcircle(A,D,E),linetype("4 4"));
label("$A$",A,dir(A));
label("$B$",B,dir(190));
label("$C$",C,dir(350));
label("$Q$",Q,dir(Q));
label("$D$",D,dir(40));
label("$E$",E,dir(220));
label("$O$",O,N);
label("$P$",P,SE);
[/asy]](http://latex.artofproblemsolving.com/e/a/6/ea63bb8060a3a3f32e8be1f76563757747546980.png)
and 
and that 
.
is the radical axis of 
is cyclic.![\[\angle ADO = \angle OBC = 90^\circ-\angle A,\]](http://latex.artofproblemsolving.com/2/e/8/2e801448ce87bfa304658ba23c434e3d0a16c483.png)
which implies 
.
,
.
and 
.
is a parallelogram. This means that ![\[2(OD^2+OE^2) = OQ^2+DE^2 = OA^2+DE^2.\]](http://latex.artofproblemsolving.com/e/1/6/e1685f08a93ea433daa5ef889018e0b893f8db66.png)
But note that if 
is the circumradius of ![\[OA^2+DE^2 = (2R'\cos A)^2 + (2R'\sin A)^2 = 4R'^2,\]](http://latex.artofproblemsolving.com/f/9/5/f95a9be841012c64062c81879fcaeb26ad219407.png)
which we know is equal to 
since ![\[AQ = \sqrt{2(OD^2+OE^2)} = \sqrt{2(3^2+7^2)} = \boxed{2\sqrt{29}}.\]](http://latex.artofproblemsolving.com/d/7/8/d78ac74f51234bc38f7a10bcdca2bfde11175211.png)
2017 CMIMC Team #9 wrote:
Circles 
and 
are externally tangent to each other. Circle 
is placed such that is internally tangent to at 
while is internally tangent to at 
. Line 
is tangent to at 
and at 
and furthermore intersects at points and 
with 
. Suppose that 
, 
, and 
. Compute the length of line segment 
.
and 
are externally tangent to each other. Circle 
is placed such that is internally tangent to at 
while is internally tangent to at 
. Line 
is tangent to at 
and at 
and furthermore intersects at points and 
with 
. Suppose that 
, 
, and 
. Compute the length of line segment 
.
is the midpoint of minor arc 
.![\[\angle XPA = \dfrac{\widehat{AX}+\widehat{MB}}2 = \dfrac{\widehat{AX}+\widehat{MA}}2 = \angle XYQ,\]](http://latex.artofproblemsolving.com/c/6/9/c694ce449d1cbac697f43ad0c913b07cd7771535.png)
whence 
is a cyclic quadrilateral. Now 
is the radical center of 
,
,
is the common internal tangent of the two circles.![[asy]
import olympiad;
size(270);
defaultpen(linewidth(0.8)+fontsize(11pt));
real r1 = 2.42, r2 = 4.56, d = sqrt(r1*r2);
pair O1 = (-1*d,r1), O2 = (d,r2), P = (-d,0), Q = (d,0);
path circ1 = circle(O1,r1), circ2 = circle(O2,r2);
draw(circ1^^circ2);
pair M = (1.34, -4.16), pt1 = 3*P - 2*M, pt2 = 3*Q - 2*M;
pair[] X = intersectionpoints(P--pt1, circ1), Y = intersectionpoints(Q--pt2, circ2);
path omega = circumcircle(M,X[1], Y[1]);
pair O = circumcenter(M,X[1],Y[1]);
draw(omega);
pair Pext = 2*P-Q, Qext = 2*Q-P;
pair[] end = intersectionpoints(Pext--Qext, omega);
pair A = end[0], B = end[1];
draw(Y[1]--X[1]^^A--B);
draw(X[1]--M--Y[1]^^O1--O2, linetype("3 3"));
pair Os = circumcenter(P,Q,X[1]);
real r = circumradius(P,Q,X[1]);
draw(circle(Os,r), linetype("4 4"));
pair T = foot(M,O1,O2);
dot(O1^^O2^^P^^Q^^M^^T^^X[1]^^Y[1]);
draw(M--T, linetype("3 3"));
label("$A$",A,1.2*dir(235));
label("$B$",B,1.2*dir(305));
label("$X$",X[1],2*dir(165));
label("$Y$",Y[1],1.5*dir(35));
label("$P$",P,1.5*dir(250));
label("$Q$",Q,1.5*dir(290));
label("$T$",T,2*dir(330));
label("$M$",M,1.5*S);
[/asy]](http://latex.artofproblemsolving.com/6/c/a/6cac7490e8b5ed011e167bbd0264f0e506610f26.png)
,
.
,![\[MA^2 = MP\cdot MX = MT^2\quad\implies\quad MT = r.\]](http://latex.artofproblemsolving.com/4/9/5/49519387ecfd5c85f91fa7378882008a22d63cf4.png)
Thus 
and 
.
with cevian 
gives us the value of ![\[
4 \cdot 5 \cdot 9 + 9(r-2)^2 = 4r^2 + 5r^2
\implies r = 6.
\]](http://latex.artofproblemsolving.com/9/6/0/960b6e6d606f1e882ed647fb357a3fc81cec6f10.png)
Finally, if we let 
and 
now, we have 
,
.![\[ x\left(x+\dfrac{14}x\right) = y\left(y+\dfrac{18}y\right) = r^2 = 36, \]](http://latex.artofproblemsolving.com/4/9/0/49053afd2293a04524559986f4c6c7224c427728.png)
so 
and 
and 
.2017 CMIMC Algebra #10 wrote:
Let denote the largest possible real number such that there exists a nonconstant polynomial with ![\[P(z^2)=P(z-c)P(z+c)\]](//latex.artofproblemsolving.com/e/d/5/ed541828dbab4324c99422e8c114708bfda726c8.png)
for all . Compute the sum of all values of 
over all nonconstant polynomials satisfying the above constraint for this .
denote the largest possible real number such that there exists a nonconstant polynomial with ![\[P(z^2)=P(z-c)P(z+c)\]](http://latex.artofproblemsolving.com/e/d/5/ed541828dbab4324c99422e8c114708bfda726c8.png)
for all . Compute the sum of all values of 
over all nonconstant polynomials satisfying the above constraint for this .
.
is a root of 
yields ![\[P((\alpha+c)^2) = P(\alpha)P(\alpha+2c) = 0,\]](http://latex.artofproblemsolving.com/d/9/5/d958ed124092beb6d0cb1a6b96ae8847ca44ffa8.png)
so that 
is a root of 
must also be a root of 
,
such that 
and such that 
is either equal to 
or 
.
is strictly increasing. To see this, recall by the Parallelogram Law, ![\[|z-c|^2+|z+c|^2 = 2(|z|^2+c^2).\]](http://latex.artofproblemsolving.com/e/8/0/e801773101553fabd74a73e7e12e330c4a0716eb.png)
It thus follows that one of 
and 
must be at least 
(else the entire sum would be too small), so we can choose 
.![\[|z|^2+c^2>|z|\quad\iff\quad \left(|z|-\frac12\right)^2 + c^2 > \frac14,\]](http://latex.artofproblemsolving.com/2/d/7/2d7cf68eb0decb9c39ed5097a82a1ed04a90a9b3.png)
which is always true for 
.
,
,
,
.
and 
are the only possible roots of 
,
Hence 
for some integer 
,![\[\sum_{n\geq 1}\left(\dfrac19+\frac14\right)^n = \sum_{n\geq 1}\left(\frac{13}{36}\right)^n = \boxed{\frac{13}{23}}.\]](http://latex.artofproblemsolving.com/f/0/3/f03af8740fb3afa666f4cfc21f021382a060cc57.png)
though....Oops this list doesn't have much variety....
This post has been edited 3 times. Last edited by djmathman, Jan 30, 2017, 2:59 AM
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