SMT Recap
by ryanbear, Apr 23, 2026, 5:06 PM
I didn't do any of the activities because my parents didn't want me to miss school
I arrived at around 11 PM and then slept
Power round:
Our team managed to finish the entire section 1 and some parts of section 2
I completely understood the concepts in section 1, which I was happy about
This power round felt way easier than the ones our team did during practice, but it still felt more like reading comprehension than math
Team round:
In the last few minutes, I managed to solve P11 for the team
I really liked that problem since it seems very simple, and has a very nice solution that manages to connect expected value with factorials through p-adic values
Alg:
I solved 1-7 at a normal pace and ran out of time to do the later problems
Discrete:
Same as alg, but I solved 1-6
I really liked p6 because I used algebra, number theory, and combinatorics knowledge to solve it, and the problem statement felt very natural
Guts:
I found the problems really bashy and I was too tired to contribute that much
We managed to get a problem wrong on the first set
Tiebreak:
The alg tiebreak was an estimation problem
The problem had an f^10(x), but I forgot the 10 was a thing and I got 111, when the answer was ~16
I think using one estimaton to determine who gets top 10 is a bad idea and they should have just used a normal tiebreaker test
Other comments:
I really liked the weather, and the region felt normal (during hmmt, there was a bunch of snow, and it felt like a completely new place)
Also, the SMT problems were way more fun than HMMT, since they were actually doable
For me, this was a really good math performance, since I solved some really hard/nice problems and contributed to my team a lot
I managed to get DHM for alg and HM for geo, and I think I got the highest indiv scores on my team
I was on a B team, and the A team did really good (top 5 overall and top 1 indiv)
I also did do any prep for SMT besides the team practices and I still did good
Overall, this year, I stopped preparing for math contests, and I have been getting better results, while other people I know have thrown (eg. AIME, SMT)
I arrived at around 11 PM and then slept
Power round:
Our team managed to finish the entire section 1 and some parts of section 2
I completely understood the concepts in section 1, which I was happy about
This power round felt way easier than the ones our team did during practice, but it still felt more like reading comprehension than math
Team round:
In the last few minutes, I managed to solve P11 for the team
I really liked that problem since it seems very simple, and has a very nice solution that manages to connect expected value with factorials through p-adic values
Alg:
I solved 1-7 at a normal pace and ran out of time to do the later problems
Discrete:
Same as alg, but I solved 1-6
I really liked p6 because I used algebra, number theory, and combinatorics knowledge to solve it, and the problem statement felt very natural
Guts:
I found the problems really bashy and I was too tired to contribute that much
We managed to get a problem wrong on the first set
Tiebreak:
The alg tiebreak was an estimation problem
The problem had an f^10(x), but I forgot the 10 was a thing and I got 111, when the answer was ~16
I think using one estimaton to determine who gets top 10 is a bad idea and they should have just used a normal tiebreaker test
Other comments:
I really liked the weather, and the region felt normal (during hmmt, there was a bunch of snow, and it felt like a completely new place)
Also, the SMT problems were way more fun than HMMT, since they were actually doable
For me, this was a really good math performance, since I solved some really hard/nice problems and contributed to my team a lot
I managed to get DHM for alg and HM for geo, and I think I got the highest indiv scores on my team
I was on a B team, and the A team did really good (top 5 overall and top 1 indiv)
I also did do any prep for SMT besides the team practices and I still did good
Overall, this year, I stopped preparing for math contests, and I have been getting better results, while other people I know have thrown (eg. AIME, SMT)
This post has been edited 1 time. Last edited by ryanbear, Apr 23, 2026, 5:49 PM
cane from from the curvature formula and why there is a cross product, so I tried to make sense of it
.
be the angle between 
and 
(the two black arrows)
by definition of sin
and 
,
is curvature.
into 
,
think that ![$I=1=[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}]=MR^2=\frac{V}{R}$](./blog_files/9a43712cd7013f73bd4f12225b452601bbc006bc.png)
and 
be the constant after taking the half derivative of 
.
to get that 
- should get ![$a^3+[]a+[]=0$](./blog_files/a7b0faa727be47ee44f0d1d897e450ab943afaaa.png)
![$a=\sqrt{\frac{-(a term)}{3}}(\sqrt[3]{k}+\sqrt[3]{\frac{1}{k}})$](./blog_files/d057f7aa262e03ab3ad0b1668b4bb4526359b8b5.png)
![$[]k+[]\frac{1}{k}+[]=0$](./blog_files/0b460ce7005e2bc46c624e979a878b8b2c8b3557.png)
through quadratic formula, then get 
(2 solutions)
'
,
,
is 
,
rectangular prism on the ground
rectangular prism right above the previous one
rectangular prism right above the previous one
rectangular prism
.
.
of them to form a 
cube.
be an integer,
be a subset of 
.
,
and 
.
is always true. Note that 
.
,
and 
.
.
.
.
works.
works. Then, 
,
also works.
.
,
such that 
and 
.
,
is repeated 
be the perpendicular distance from the center of a cube to the edge. Note that the volume of the cube is 
.
.
,
faces and that an 
faces.
is divided by 
.
.
and 
.
and 
.
is equal to 
.
rows and 
columns. There are hidden monsters in 
of the cells. Initially, Turbo does not know where any of the monsters are, but he knows that there is exactly one monster in each row except the first row and the last row, and that each column contains at most one monster.
![\[k = \frac{x^2-a}{x^2-y^2}.\]](./blog_files/c223ccbc272d07c19632c5b8883571023e47a395.png)
Let 
with 
,
be the set of positive integers for which the equation admits a solution in 
.
.
be a positive integer and 
be a positive integer coprime to 
,
,
Find, in terms of 
,
is divisible by 
.
,
,
.
.
and 
are between 
,
,
be the common element.
,
.
,
,
,
,
.
but for 
,
(i did not know this)
is the remainder upon dividing 
by 
.
can be expressed as 
.
.
be 
by the remainder theorem
and when 
this becomes 
,
are arranged in the coordinate plane so that they form a regular pentagon. Point 
follows point 
and point 
follows 
.
be the distance between two consecutive real numbers 
form a regular polygon, then for the next 
also works, so by induction, all times work.
![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](./blog_files/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
.
.
and 
.
to 
,
,
,
be triangle with incenter 
in the interior of the triangle satisfies ![\[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\]](./blog_files/d622094b95a1f317063cf2f02a8b66b31751638f.png)
Show that 
,
.
is cyclic.
be the point where the line 
intersects the circumcircle of 
,
![https://cdn.discordapp.com/attachments/1140707511859679364/1187262830341259314/image.png?ex=65963fad&is=6583caad&hm=b5773143920794a5c7265a6dccc4e159b04ccacd6cb5912269406e6bcba6cdec&]](./blog_files/2e77ca2269b0255a4465b0811597261e822f6131.png)
from 
,
.
,
.
is 
.
.
becomes 
,
this becomes 
,
or 
for more accuracy, where 
and 
,
![$\frac{2(ab+bc+ca)^3}{(abc)^2}+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)+1 \ge \frac{2(3\sqrt[3]{a^2b^2c^2})^3}{(abc)^2}+9+1 \ge 64$](./blog_files/3eca66a55ca46bbf4e44233f8aa352efed70ff81.png)
by cauchy and AM-GM
be the roots of the equation 
.
such that
are consecutive primes and a number between 
dos not work, then anything larger than the number also does not work because the factorial is multiplied but the prime products stay the same.
results in 
working. But 
also does not work.
because 
,
are different mod 
,
is an odd that is divisible by 
,
because two primes only sum to an odd if one is 
for all positive 
,
.
when 
is prime, which is less than 
,
,
.
.
.
,
is a multiple of 
do not work
repeats every 
repeats every 
by euler totent, so 
repeats every 
to get that 
repeats every 
by euler totent, so 
mod 
,
(impossible)
,
(possible)
(because 
)
to get that 
works
is continuous, solve 
over the reals
--> 
is linear and 
is linear
results in 
or 
April 2026
March 2026