It's been a while since I've posted math on here
by djmathman, Jul 19, 2024, 1:21 AM
so have a random collection of problems I've solved over the past 
months, where 
.
CMIMC 2024 Algebra 9 (Robert Trosten). Let
be the non-negative rational numbers, 
such that 
, 
for 
, and 
Define a sequence 
of non-negative integers recursively via 
for every 
. Find 
CMIMC 2024 Algebra 10 (Connor Gordon). There exists a unique pair of polynomials
such that
Compute 
.
Romania District Olympiad 2024 Grade 12 Problem 4. Let
be a differentiable function, with a continous derivative. Given that 
and 
for every 
prove that![\[\frac{1}{n+1}\int_0^af(t)^{2n+1}\mathrm{d}t\leqslant\left(\int_0^af(t)^n\mathrm{d}t\right)^2,\]](//latex.artofproblemsolving.com/5/f/1/5f19be19c27318882cecb0e6e013cf3e2678f0a9.png)
for any positive integer 
and real number 
OTIS Mock AIME Problem 15 (Wilbert Chu). A parabola in the Cartesian plane is tangent to the
-axis at 
and to the 
-axis at 
. Find the sum of the coordinates of the vertex of the parabola.
OMMC PoTM March 2022 (Evan Chang). Define acute triangle
with circumcircle 
Let 
be the midpoint of minor arc 
in 
and let 
be the reflection of over 
If the circle with diameter is tangent to the external angle bisector of 
at 
show 
BAMO 2010 Problem 4. Acute triangle has 
. Point 
lies in the interior of triangle so that 
and 
. Point 
is the reflection of 
across line 
, and point 
is the reflection of 
across line 
. Prove that lines 
and 
are perpendicular.
Naoki Sato. In triangle, sides and are extended to and , respectively, so that 
is parallel to and tangent to the 
-excircle. Construct the circle that passes through and , and is tangent to the -excircle. Prove that this circle is also tangent to the incircle of triangle .
![[asy]
real area(pair A, pair B, pair C) {
return(abs((xpart(A)*ypart(B) + xpart(B)*ypart(C) + xpart(C)*ypart(A) - ypart(A)*xpart(B) - ypart(B)*xpart(C) - ypart(C)*xpart(A)))/2);
};
pair excentre(pair A, pair B, pair C) {
return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
real exradius(pair A, pair B, pair C) {
return(2*area(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
path excircle(pair A, pair B, pair C) {
return(Circle(excentre(A,B,C),exradius(A,B,C)));
};
unitsize (0.28 cm);
pair A, B, C, D, E, P, T, X;
A = (0,0);
B = (-1,-4);
C = (5,-4);
D = interp(A,B,exradius(A,B,C)/inradius(A,B,C));
E = interp(A,C,exradius(A,B,C)/inradius(A,B,C));
X = (-9.19135,-14.6049);
T = (incenter(A,D,E) + reflect(D,E)*(incenter(A,D,E)))/2;
P = reflect(X,incenter(A,D,E))*(T);
draw(incircle(A,B,C),red);
draw(excircle(A,B,C),red);
draw(circumcircle(D,E,P),blue);
draw(A--D--E--cycle);
draw(B--C);
label("$A$", A, N);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
[/asy]](//latex.artofproblemsolving.com/7/3/b/73be5d3f064c85d2bfe424bec10c98f69e6f6da9.png)
months, where 
.CMIMC 2024 Algebra 9 (Robert Trosten). Let
be the non-negative rational numbers, 
such that 
, 
for 
, and 
Define a sequence 
of non-negative integers recursively via 
for every 
. Find 
![\[
[x_0;x_1,x_2,\ldots, x_n] := x_0 + \cfrac{1}{x_1 + \cfrac{1}{x_2 + \cfrac{1}{\cdots + \cfrac{1}{x_n}}}}
\]](http://latex.artofproblemsolving.com/0/7/b/07b589ccfcbfca216daa9fe0288a4270ac9d07f3.png)
for any real numbers 
.
.![$q\sim [a_0;a_1,a_2,\ldots, a_n]$](http://latex.artofproblemsolving.com/9/d/a/9dad8b74b7b2e832dc4267954207326c07981fff.png)
is the simple continued fraction representation of a rational number 
,![\[
f(q) = a_0 + a_1 + a_2 + \cdots + a_n.
\]](http://latex.artofproblemsolving.com/a/d/1/ad1d3439077e871af7f05df0dde30352fce0440d.png)
Proof. Induction. The base case holds by inspection, and for the inductive step, remark that![\begin{align*}
f([a_0;a_1,a_2,\ldots, a_n]) &= a_0 + f([0;a_1,a_2,\ldots, a_n]) \\
&= a_0 + f([a_1;a_2,\ldots, a_n]) \\
&\stackrel{\textrm{IH}}=a_0 + a_1 + \cdots + a_n. \quad\square
\end{align*}](http://latex.artofproblemsolving.com/e/d/5/ed52335bf9cb0f091367ec622f71da2aac02221f.png)
![$q = [a_0;-a_1,-a_2,\ldots, -a_n]$](http://latex.artofproblemsolving.com/d/b/1/db1aeefe3c3e5348abaee63575c92d3e2adfb40c.png)
,
,![\[
f(q) = a_0 + a_1 + \cdots + a_n - n.
\]](http://latex.artofproblemsolving.com/b/2/1/b214a5f947e5a15ff66c5c5a47d87cdadcb169d9.png)
![\[
a - \frac 1q = a - 1 + \frac{q-1}q = a - 1 + \cfrac{1}{\frac{q}{q-1}} = a - 1 + \cfrac{1}{1 + \cfrac{1}{q-1}},
\]](http://latex.artofproblemsolving.com/a/f/7/af75e83c81abb9ea00551a8dc1c0681de485c4e1.png)
so 
.
from the overall sum when 
for each positive integer 
;
and 
.
is 
,
and 
satisfying 
and 
.
has characteristic polynomial with roots 
and 
,
.![\[
Q_n = 34 Q_{n-1} - Q_{n-2}\text{ for every }n\geq 2,
\]](http://latex.artofproblemsolving.com/f/4/c/f4cd879461ea3c10aac7be5b061c5528dfcdb1ea.png)
or![\[
\frac{Q_{n}}{Q_{n-1}} = 34 - \frac{Q_{n-2}}{Q_{n-1}}.
\]](http://latex.artofproblemsolving.com/5/f/0/5f0e4fd6ca5f27f004feb90362108439181c9771.png)
It follows from induction and 
that![\[
\frac{Q_6}{Q_5} = [34;-34,-34,-34,-34],
\]](http://latex.artofproblemsolving.com/5/c/4/5c48d00b817766ca497d36c6b7ff1a95524a044a.png)
so 
.CMIMC 2024 Algebra 10 (Connor Gordon). There exists a unique pair of polynomials
such that
Compute 
.
and 
.
and 
be the roots of 
,
and 
analogously. Observe that![\[
P(Q(x)) = \pm(Q(x) - r_P)(Q(x) - r_Q)
\]](http://latex.artofproblemsolving.com/c/a/b/cab19121e6f91dca559585ac4d6ff0d8716eaeb4.png)
is the product of two factors, each of whose sum of roots is 
.
.
.
and 
.![\[
P(x) = \pm(x^2 - 4x + b)\quad\text{and}\quad Q(x) = \pm(x^2 - 5x + d)
\]](http://latex.artofproblemsolving.com/8/a/8/8a8fa51cad7e1da4240c6b081cf72d45847347c8.png)
for some independent choices of signs.
and 
,
,
.![\[
P(0)(0^2 - 6\cdot 0 + 7) = P(5)(5^2 - 6\cdot 5 + 7).
\]](http://latex.artofproblemsolving.com/e/0/1/e014ecb505e45384d0d7eb2d0f28d156d6897464.png)
Regardless of the leading coefficient of 
,
.
,
and![\[
Q(0)(0^2 - 3\cdot 0 - 2) = Q(4)(4^2 - 3\cdot 4 - 2).
\]](http://latex.artofproblemsolving.com/9/e/b/9eb73713432fb448236f791d9fd2e0d130cf467a.png)
Regardless of the leading coefficient of 
,
.
and 
.
,
,
.
,
,
.![\[
P(10) + Q(-10) = (10^2 - 4\cdot 10 + 2) - [(-10)^2 - 5\cdot(-10) + 2] = \boxed{-90}.
\]](http://latex.artofproblemsolving.com/1/a/1/1a1dc6dd3e0275664b5c859e27c698d22a5ef159.png)
Romania District Olympiad 2024 Grade 12 Problem 4. Let
be a differentiable function, with a continous derivative. Given that 
and 
for every 
prove that![\[\frac{1}{n+1}\int_0^af(t)^{2n+1}\mathrm{d}t\leqslant\left(\int_0^af(t)^n\mathrm{d}t\right)^2,\]](http://latex.artofproblemsolving.com/5/f/1/5f19be19c27318882cecb0e6e013cf3e2678f0a9.png)
for any positive integer 
and real number 
.
because 
for all 
be arbitrary. Starting with the bound 
,
to ![\[ \frac{f(u)^{n+1}}{n+1} \leq \int_0^u f(t)^n\,dt. \]](http://latex.artofproblemsolving.com/c/0/f/c0fc7b01fc565b60f65e589469f820644a2e498b.png)
Multiplying both sides by 
gives![\[ \frac{f(u)^{2n+1}}{n+1} \leq f(u)^n\int_0^u f(t)^n\,dt. \]](http://latex.artofproblemsolving.com/0/6/7/06734e97a43a4f80b46bf69bd916b6b781bec90f.png)
But, by FTC Part 1, the left hand side is the derivative of 
evaluated at 
,
evaluated at 
gives the desired inequality improved by a factor of 
,![\[ \frac{1}{n+1}\int_0^a t^{2n+1}\,dt = \frac12 a^{2(n+1)} = \frac12\left(a^{n+1}\right)^2 = \frac12\left(\int_0^a t^n\,dt\right)^2. \]](http://latex.artofproblemsolving.com/e/8/6/e866ec7cda141a8642327b410e286868417581ec.png)
OTIS Mock AIME Problem 15 (Wilbert Chu). A parabola in the Cartesian plane is tangent to the
-axis at 
and to the 
-axis at 
. Find the sum of the coordinates of the vertex of the parabola.
and 
be the tangency points of the parabola with the 
be the origin, and let 
be the focus and vertex, respectively, of the parabola. Furthermore, let 
and 
be the projections of 
of the parabola passes through 
,![\[
\frac{QV}{VX} = \frac{YV}{VP} = \frac{FQ}{FP}.
\]](http://latex.artofproblemsolving.com/7/4/6/746a900e59384cbae52829495a7ac4279c7072d3.png)
![[asy]
size(180);
defaultpen(linewidth(0.7));
pair O = origin, X = (2,0), Y = (-2,0), F = 2 * dir(58), V =(F.x,F.y/2), T = (F.x,0);
pair AA = rotate(90,F) * O, P = extension(F,AA,X,(2,1)), Q = extension(F,AA,Y,(-2,1));
real f (real x)
{
real c = (P.y-V.y)/(P.x-V.x)^2;
return c*(x - V.x)^2 + V.y;
}
draw(graph(f, -2.3,2.3),gray(0.6),Arrows(size=7));
draw(P--Q--O--cycle^^rightanglemark(P,O,Q,5),rgb(0.9,0.1,0.1));
draw(O--F^^rightanglemark(O,F,P,5),rgb(0.95,0.35,0.25));
draw(X--Y, rgb(0.1,0.1,0.7));
draw(Q--Y^^F--T^^P--X^^rightanglemark(Q,Y,O,5)^^rightanglemark(F,T,X,5)^^rightanglemark(P,X,O,5), rgb(0.6,0.1,0.6));
dot("$F$",F,NE,linewidth(3.3));
dot("$P$",P,SE,linewidth(3.3));
dot("$Q$",Q,NE,linewidth(3.3));
dot("$X$",X,S,linewidth(3.3));
dot("$Y$",Y,S,linewidth(3.3));
dot("$T$",T,S,linewidth(3.3));
dot("$V$",V,SW,linewidth(3.3));
dot("$O$",O,S,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/0/1/3/0137bd9deb49c89e0a049f1e4bb9f0d30cacd4c0.png)
is the perpendicular bisector of 
,
is the perpendicular bisector of 
.
.
,
too, and that 
.
,
.
.
and 
.
the foot of the perpendicular from 
;
.
is a weighted average of the lengths 
and 
;![\[
FT = \frac{x}{x+y}\cdot QY + \frac{y}{x+y}\cdot PX = \frac{2xy}{x+y}.
\]](http://latex.artofproblemsolving.com/7/9/d/79d1665171b6c7d98a929090df64d0be62215a3f.png)
Thus 
.
and 
,
.
,![\[
V = \frac{9}{10}X + \frac{1}{10}Q = \left(\frac{81}{100},\frac{3}{100}\right).
\]](http://latex.artofproblemsolving.com/d/b/3/db3ada37e0910a1a2bea15862728c18393d2c572.png)
Hence the sum of the coordinates is 
.OMMC PoTM March 2022 (Evan Chang). Define acute triangle
with circumcircle 
Let 
be the midpoint of minor arc 
in 
and let 
be the reflection of over 
If the circle with diameter is tangent to the external angle bisector of 
at 
show 
be the midpoint of 
and 
and 
be the circle with diameter 
and 
be the intersection point of 
and 
,
is the external angle bisector and 
is the internal angle bisector, ergo 
.
.
and 
have the same length.![\[
TA\cdot TQ = TB\cdot TC = TP^2.
\]](http://latex.artofproblemsolving.com/0/8/5/085cadc1afe96cb4b9672feefdc39d6bc6999276.png)
Additionally, by Claim 1, triangles 
and 
are similar, so 
.
.
,
.
.BAMO 2010 Problem 4. Acute triangle
has 
. Point 
lies in the interior of triangle so that 
and 
. Point 
is the reflection of 
across line 
, and point 
is the reflection of 
across line 
. Prove that lines 
and 
are perpendicular.
and 
.
from an angle chase, hence 
is cyclic. Because 
,
as well. Rephrasing the problem from the perpsective of 
,
.
and 
such that 
.
.
and
The claim follows by the Perpendicularity Lemma.
,Naoki Sato. In triangle
, sides and are extended to and , respectively, so that 
is parallel to and tangent to the 
-excircle. Construct the circle that passes through and , and is tangent to the -excircle. Prove that this circle is also tangent to the incircle of triangle .![[asy]
real area(pair A, pair B, pair C) {
return(abs((xpart(A)*ypart(B) + xpart(B)*ypart(C) + xpart(C)*ypart(A) - ypart(A)*xpart(B) - ypart(B)*xpart(C) - ypart(C)*xpart(A)))/2);
};
pair excentre(pair A, pair B, pair C) {
return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
real exradius(pair A, pair B, pair C) {
return(2*area(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
path excircle(pair A, pair B, pair C) {
return(Circle(excentre(A,B,C),exradius(A,B,C)));
};
unitsize (0.28 cm);
pair A, B, C, D, E, P, T, X;
A = (0,0);
B = (-1,-4);
C = (5,-4);
D = interp(A,B,exradius(A,B,C)/inradius(A,B,C));
E = interp(A,C,exradius(A,B,C)/inradius(A,B,C));
X = (-9.19135,-14.6049);
T = (incenter(A,D,E) + reflect(D,E)*(incenter(A,D,E)))/2;
P = reflect(X,incenter(A,D,E))*(T);
draw(incircle(A,B,C),red);
draw(excircle(A,B,C),red);
draw(circumcircle(D,E,P),blue);
draw(A--D--E--cycle);
draw(B--C);
label("$A$", A, N);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
[/asy]](http://latex.artofproblemsolving.com/7/3/b/73be5d3f064c85d2bfe424bec10c98f69e6f6da9.png)
![[asy]
defaultpen(fontsize(10pt));
real area(pair A, pair B, pair C) {
return(abs((xpart(A)*ypart(B) + xpart(B)*ypart(C) + xpart(C)*ypart(A) - ypart(A)*xpart(B) - ypart(B)*xpart(C) - ypart(C)*xpart(A)))/2);
};
pair excentre(pair A, pair B, pair C) {
return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
real exradius(pair A, pair B, pair C) {
return(2*area(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
path excircle(pair A, pair B, pair C) {
return(Circle(excentre(A,B,C),exradius(A,B,C)));
};
unitsize (0.3 cm);
pair A, B, C, D, E, I, Ia, U, V, P, Q, R, S, T;
A = (0,0);
B = (-1,-4);
C = (5,-4);
D = interp(A,B,exradius(A,B,C)/inradius(A,B,C));
E = interp(A,C,exradius(A,B,C)/inradius(A,B,C));
I = incenter(A,B,C); Ia = excentre(A,B,C);
T = foot(I, B,C); U = foot(I,A,B); V = foot(I,A,C); S = foot(Ia, B, C); P = foot(Ia,A,D); Q = foot(Ia,A,E); R = foot(Ia,D,E);
draw(incircle(A,B,C),red);
draw(excircle(A,B,C),red);
draw(A--D--E--cycle);
draw(B--C);
label("$A$", A, N);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
label("$U$",U,NW);
label("$V$",V,NE);
label("$S$",S,dir(270));
label("$T$",T, N);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$R$",R,dir(270));
[/asy]](http://latex.artofproblemsolving.com/0/d/c/0dcf24d5798ec2381580050753191675c75b8e61.png)
.
its 
and 
be the tangency points of 
,
,
,
.![\begin{align*}
BC\cdot TS + BT\cdot EV &= a(b-c) + (s-b)\left[a + \frac{(s-c)s}{s-a}\right]\\
&= a(b-c+s-b) + \frac{s(s-b)(s-c)}{s-a}\\
&= a(s-c) + (s-c)\cdot\frac{s(s-b)}{s-a} = CT\cdot DU.
\end{align*}](http://latex.artofproblemsolving.com/6/3/9/639556b1c44b845439b970df20e83432a1d76479.png)
Scaling up by a factor of 
yields the equivalent equality![\[
DE\cdot TS + DR\cdot EV = ER\cdot DU.
\]](http://latex.artofproblemsolving.com/f/e/5/fe521d06e1828d35816824f80267aff0f212f309.png)
It follows by the converse to Casey's Theorem that there exists a circle passing through the point circles 
and externally tangent to This post has been edited 2 times. Last edited by djmathman, Jul 19, 2024, 6:19 PM
April 2025
November 2024