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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
inequalities of elements in set
toanrathay   0
5 minutes ago
Let \( m \) be a positive integer such that \( m \geq 4 \), and let the set
\[
A = \{a_1, a_2, a_3, \ldots, a_m\}
\]consist of distinct positive integers not exceeding 2025. Suppose that for every \( a, b \in A \), with \( a \ne b \), if \( a + b \leq 2025 \), then \( a + b \in A \) as well. Prove that

\[
\frac{a_1 + a_2 + a_3 + \cdots + a_m}{m} \geq 1013.
\]
0 replies
toanrathay
5 minutes ago
0 replies
nice geo
Melid   1
N 31 minutes ago by Melid
Source: 2025 Japan Junior MO preliminary P9
Let ABCD be a cyclic quadrilateral, which is AB=7 and BC=6. Let E be a point on segment CD so that BE=9. Line BE and AD intersect at F. Suppose that A, D, and F lie in order. If AF=11 and DF:DE=7:6, find the length of segment CD.
1 reply
Melid
36 minutes ago
Melid
31 minutes ago
product of all integers of form i^3+1 is a perfect square
AlastorMoody   3
N 37 minutes ago by Assassino9931
Source: Balkan MO ShortList 2009 N3
Determine all integers $1 \le m, 1 \le n \le 2009$, for which
\begin{align*} \prod_{i=1}^n \left( i^3 +1 \right) = m^2 \end{align*}
3 replies
AlastorMoody
Apr 6, 2020
Assassino9931
37 minutes ago
IMO ShortList 1998, combinatorics theory problem 1
orl   44
N 44 minutes ago by YaoAOPS
Source: IMO ShortList 1998, combinatorics theory problem 1
A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)
44 replies
orl
Oct 22, 2004
YaoAOPS
44 minutes ago
A game optimization on a graph
Assassino9931   2
N an hour ago by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[
  \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
  \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bobby has a winning strategy.
2 replies
Assassino9931
Apr 8, 2025
dgrozev
an hour ago
Composite sum
rohitsingh0812   39
N an hour ago by YaoAOPS
Source: INDIA IMOTC-2006 TST1 PROBLEM-2; IMO Shortlist 2005 problem N3
Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers and let $ S = a+b+c+d+e+f$.
Suppose that the number $ S$ divides $ abc+def$ and $ ab+bc+ca-de-ef-df$. Prove that $ S$ is composite.
39 replies
rohitsingh0812
Jun 3, 2006
YaoAOPS
an hour ago
Problem 1
SpectralS   145
N an hour ago by IndexLibrorumProhibitorum
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
145 replies
SpectralS
Jul 10, 2012
IndexLibrorumProhibitorum
an hour ago
Help me please
sealight2107   0
an hour ago
Let $m,n,p,q$ be positive reals such that $m+n+p+q+\frac{1}{mnpq} = 18$. Find the minimum and maximum value of $m,n,p,q$
0 replies
sealight2107
an hour ago
0 replies
Rectangular line segments in russia
egxa   2
N an hour ago by mohsen
Source: All Russian 2025 9.1
Several line segments parallel to the sides of a rectangular sheet of paper were drawn on it. These segments divided the sheet into several rectangles, inside of which there are no drawn lines. Petya wants to draw one diagonal in each of the rectangles, dividing it into two triangles, and color each triangle either black or white. Is it always possible to do this in such a way that no two triangles of the same color share a segment of their boundary?
2 replies
egxa
Apr 18, 2025
mohsen
an hour ago
(help urgent) Classic Geo Problem / Angle Chasing?
orangesyrup   4
N an hour ago by Royal_mhyasd
Source: own
In the given figure, ABC is an isosceles triangle with AB = AC and ∠BAC = 78°. Point D is chosen inside the triangle such that AD=DC. Find the measure of angle X (∠BDC).

ps: see the attachment for figure
4 replies
orangesyrup
2 hours ago
Royal_mhyasd
an hour ago
Interesting combinatoric problem on rectangles
jaydenkaka   0
an hour ago
Source: Own
Define act <Castle> as following:
For rectangle with dimensions i * j, doing <Castle> means to change its dimensions to (i+p) * (j+q) where p,q is a natural number smaller than 3.

Define 1*1 rectangle as "C0" rectangle, and define "Cn" ("n" is a natural number) as a rectangle that can be created with "n" <Castle>s.
Plus, there is a constraint for "Cn" rectangle. The constraint is that "Cn" rectangle's area must be bigger than n^2 and be same or smaller than (n+1)^2. (n^2 < Area =< (n+1)^2)

Let all "C20" rectangle's area's sum be A, and let all "C20" rectangles perimeter's sum be B.
What is A-B?
0 replies
jaydenkaka
an hour ago
0 replies
Complex Numbers Question
franklin2013   2
N 2 hours ago by Xx_BABAI_xX
Hello everyone! This is one of my favorite complex numbers questions. Have fun!

$f(z)=z^{720}-z^{120}$. How many complex numbers $z$ are there such that $|z|=1$ and $f(z)$ is an integer.

Hint
2 replies
franklin2013
Apr 20, 2025
Xx_BABAI_xX
2 hours ago
Inequalities
sqing   17
N 2 hours ago by sqing
Let $ a,b,c> 0 $ and $ ab+bc+ca\leq  3abc . $ Prove that
$$ a+ b^2+c\leq a^2+ b^3+c^2 $$$$ a+ b^{11}+c\leq a^2+ b^{12}+c^2 $$
17 replies
sqing
Yesterday at 1:54 PM
sqing
2 hours ago
How many ways can we indistribute n different marbles into 6 identical boxes
Taiharward   9
N 3 hours ago by mathprodigy2011
How many ways can we distribute n indifferent marbles into 6 identical boxes and one jar?
9 replies
Taiharward
Today at 2:14 AM
mathprodigy2011
3 hours ago
[MAT 2022 #9] Add Me
ApraTrip   14
N Jul 25, 2023 by ike.chen
Let $\triangle ABC$ be a scalene triangle with incenter $I$, and let the midpoint of side $\overline{BC}$ be $M$. Suppose that the feet of the altitudes from $I$ to $\overline{AC}$ and $\overline{AB}$ are $E$ and $F$ respectively, and suppose that there exists a point $D$ on $\overline{EF}$ such that $ADMI$ is a parallelogram. If $AB+AC = 24$, $BC = 14$, and the area of $\triangle ABC$ can be expressed as $m\sqrt{n}$ where $m$ and $n$ are positive integers such that $n$ is squarefree, find $m+n$.

Aprameya Tripathy
14 replies
ApraTrip
Jul 23, 2022
ike.chen
Jul 25, 2023
[MAT 2022 #9] Add Me
G H J
G H BBookmark kLocked kLocked NReply
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ApraTrip
852 posts
#1 • 1 Y
Y by Mango247
Let $\triangle ABC$ be a scalene triangle with incenter $I$, and let the midpoint of side $\overline{BC}$ be $M$. Suppose that the feet of the altitudes from $I$ to $\overline{AC}$ and $\overline{AB}$ are $E$ and $F$ respectively, and suppose that there exists a point $D$ on $\overline{EF}$ such that $ADMI$ is a parallelogram. If $AB+AC = 24$, $BC = 14$, and the area of $\triangle ABC$ can be expressed as $m\sqrt{n}$ where $m$ and $n$ are positive integers such that $n$ is squarefree, find $m+n$.

Aprameya Tripathy
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Taco12
1757 posts
#2 • 1 Y
Y by megarnie
bary ftw $                $
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bissue
302 posts
#3 • 12 Y
Y by megarnie, I-_-I, fuzimiao2013, ApraTrip, centslordm, ihatemath123, insertionsort, Bradygho, mathleticguyyy, rayfish, Mango247, sargamsujit
:love: :love: :love: :love: :love:

PART I. Standard Observations

First make a few computational observations:
  • The semiperimeter is $s = \tfrac{AB + AC + BC}{2} = \tfrac{24 + 14}{2} = 19$.
  • The lengths of $AE = AF$ are $s - BC = 19 - 14 = 5$.
We can also make the following geometric observation:
Claim wrote:
Lines $AI$ and $MD$ are perpendicular to $EF$. In particular, $D$ is necessarily the foot of the altitude from $M$ onto $EF$.
Proof: Observe that since $\overline{AI}$ bisects $\angle A$ in $\triangle ABC$, it also bisects $\angle A$ in $\triangle AEF$. However, since $\triangle AEF$ is isosceles, the angle bisector $\overline{AI}$ must also be an altitude, so $AI \perp EF$.

Consequently, since $ADMI$ is a parallelogram, $MD \perp EF$ as well. $\square$

Part II. wait what how did that just work

In order to learn more about $\triangle ABC$, we'll focus primarily on the condition that $AI = MD$ (which follows from $AIMD$ being a parallelogram).

Suppose $X$ and $Y$ are the feet of the altitudes from $B$ and $C$ onto line $EF$. Observe that:
$$MD = \dfrac{BX + CY}{2}.$$This follows from the fact that $M$ is the midpoint of $\overline{BC}$. We'll now compute $MD$ by computing $BX$ and $CY$ separately

Suppose that $AI$ has length $x$. Observe that:
$$\triangle BXF \sim \triangle AFI \hbox{ and } \triangle AEI \sim \triangle CYE.$$This follows from the fact that $BX \parallel AI \parallel CY$, as all three are altitudes onto $EF$. These similarities give these length ratios:
$$\dfrac{BX}{BF} = \dfrac{AF}{AI} = \dfrac{5}{AI} \hbox{ and } \dfrac{CY}{CE} = \dfrac{AE}{AI} = \dfrac{5}{AI}.$$Therefore we may extract:
$$BX = \dfrac{5}{AI} \times BF \hbox{ and } CY = \dfrac{5}{AI} \times CE.$$So then:
$$MD = \dfrac{BX + CY}{2} = \dfrac{5}{2 \times AI} \times (BF + CE).$$Except notice that incircle tangencies form equal lengths! In particular, if we denote the third altitude from $I$ onto $\overline{BC}$ as $G$, then $BF = BG$ and $CE = CG$, meaning:
$$MD = \dfrac{5}{2 \times AI} \times (BG + CG) = \dfrac{5}{2 \times AI} \times BC = \dfrac{35}{AI}.$$But we know that $MD$ should actually equal $AI$, meaning that:
$$AI = \dfrac{35}{AI} ~ \Longrightarrow ~ \boxed{AI = \sqrt{35}.}$$
Part III. Conclusion

Now that we know the length of $AI$, we may compute the inradius by looking at right triangle $AFI$:
$$AF^2 + FI^2 = AI^2 ~ \Longrightarrow ~ 5^2 + FI^2 = (\sqrt{35})^2 ~ \Longrightarrow FI = \hbox{ inradius } = \sqrt{10}.$$So the area is $rs = \boxed{19\sqrt{10}}$, giving an answer of $\boxed{29.}$ $\blacksquare$
This post has been edited 2 times. Last edited by bissue, Jul 24, 2022, 12:47 AM
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samrocksnature
8791 posts
#4 • 2 Y
Y by Arrowhead575, Taco12
WAIT WHAT THE ACTUAL HECK I GUESSED 10\sqrt{19} --> 29 LMHO

but i don't think i submitted by 4:00 rip

Fakesolve Solution
This post has been edited 1 time. Last edited by samrocksnature, Jul 24, 2022, 12:15 AM
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ihatemath123
3445 posts
#5
Y by
bissue here writing five paragraph essays on math problems
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Geometry285
902 posts
#6
Y by
NOOO I spent $20$ minutes on this problem and got the same observation but started some strange bash and got a weird radical :(
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squareman
966 posts
#7 • 11 Y
Y by Arrowhead575, ApraTrip, centslordm, megarnie, insertionsort, I-_-I, samrocksnature, mathleticguyyy, rayfish, ike.chen, sargamsujit
Here's a slicker approach using Iran Lemma.

Let $BI,CI$ intersect $EF$ at $X,Y.$ So $X,Y$ lie on $(BC)$ by Iran. Easy angle chase gives $\angle XMY = \angle BAC$ and also notice $DM \perp XY$ so $D$ is the midpoint of $XY.$ Then by some similar triangle ratios

$$\frac{DM}{7} = \frac{DM}{MY}= \frac{AF}{AI} = \frac{5}{AI} $$
so $AI = DM = \sqrt{35},$ rest just computation. $\blacksquare$
Attachments:
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Arrowhead575
2281 posts
#8
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:o $                 $
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euler12345
488 posts
#9
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But how many people would have known that lemma??
:noo:
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ilikemath40
500 posts
#10
Y by
euler12345 wrote:
But how many people would have known that lemma??
:noo:

I think its somewhat well known in oly geo.
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franzliszt
23531 posts
#11 • 3 Y
Y by samrocksnature, Taco12, Mango247
Congrats on this nice problem @ApraTrip! :)

First, note that $s=\frac{a+b+c}2=19, a=14,b=24-c,c=c$. Now employ barycentrics on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Clearly $M=(0,\tfrac12,\tfrac12), E=(s-c:0:s-a)=(19-c:0:5),F=(s-b:s-a:0)=(-5+c:5:0),I=(a:b:c)=(14:24-c:c)$. Since $ADMI$ is a parallelogram, $A+M=D+I \Rightarrow D=(24:-5+c:19-c).$ However, we also know that $D,E,F$ are colinear so we must have $$\begin{vmatrix}-5+c&5&0\\ 19-c&0&5\\ 24&-5+c&19-c\end{vmatrix}=0 \iff c^2-24c+133=0.$$By the quadratic formula, $c=12\pm \sqrt{11}$ and WLOG take $12+\sqrt{11}$. Then $b=24-c=12-\sqrt{11}$. Now, by Heron's $$[ABC]=\sqrt{19(19-14)(19-12+\sqrt{11})(19-12-\sqrt{11})}=\sqrt{19(5)(7^2-11)}=19\sqrt{10}.$$
This post has been edited 4 times. Last edited by franzliszt, Jul 24, 2022, 2:23 AM
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MathIsFun286
145 posts
#12
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We have $BC=BE+CF=14$. Since $AB+AC=24$ and $BE+CF=14$, we have $AE=AF=5$ and $s=19$. Proceed as in post 3 by dropping altitudes from $B$ and $C$ on to $\overline{EF}$. $BX+CY=2AI$. We next note that because of parallel lines $AI$, $MD$, $BX$, and $CY$, we have equal angles. So $BX=BE\cos A/2$ and $CY=CF\cos A/2$. And because $\cos A/2 = 5/AI$, we have $2AI=70/AI$, or $AI=\sqrt{35}$. Thus, $r=\sqrt{10}$, so $[ABC]=19r=19\sqrt{10}$.
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HamstPan38825
8857 posts
#13 • 1 Y
Y by sargamsujit
The idea is to use the condition $AI = DM$, as $D$ must be the foot of the altitude from $M$ to $\overline{EF}$. Here, notice that $$AI = \frac{AF}{\cos \frac A2} = \frac{AB+AC-BC}{2\cos \frac A2} = \frac 5{\cos \frac A2},$$and $DM$ is the average of the distances from $B, C$ to $\overline{EF}$, which is $$\frac{BF+CE}2 \cdot \cos \frac A2 = 7 \cos \frac A2.$$Thus equating the two, $\cos^2 \frac A2 = \frac 57$, and $\cos A = \frac 37$. To extract $AB \cdot AC$, we can use the Law of Cosines to get $$AB^2+AC^2-2 \cdot AB \cdot AC \cdot \frac 37 = 196 \implies (AB+AC)^2-\frac{20}7 \cdot AB \cdot AC = 196.$$Thus $AB \cdot AC = 133$, and the area is $$[ABC] = 133 \cdot \frac{2\sqrt{10}}7 \cdot \frac 12 = 19\sqrt{10}.$$
This post has been edited 1 time. Last edited by HamstPan38825, Dec 23, 2022, 8:15 PM
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Taco12
1757 posts
#14 • 5 Y
Y by samrocksnature, megarnie, Mango247, Mango247, Mango247
Solution from OTIS:

Apply barycentric coordinates on $\triangle ABC$. Note that $E=(19-c:0:5), F=(19-b:5:0)$, so $EF$ has equation $5z(19-c)+5y(19-b)=25x$. By Parallelogram Lemma, $D=\left(\frac{24}{38}, \frac{19-b}{38}, \frac{19-c}{38}\right)$ since $M=(0:1:1)$. Plugging in $D$ into the equation for $EF$ gives $b,c = 12-\sqrt{11}, 12+\sqrt{11}$ in some order. Plugging into Heron's now gives $19\sqrt{10}$.
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ike.chen
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#15
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Squareman's solution is absolutely unreal; here is a more reasonable approach.


Walk-Through:
  • Let $AI$ meet $EF$ and $(ABC)$ again at $N$ and $X$ respectively. We define $P$ and $Q$ as the projections of $X$ onto $AB$ and $AC$ respectively.
  • By Simson Lines, $M \in PQ$. Also, symmetry implies $EF \perp \overline{ANIX} \perp \overline{MPQ}$.
  • The parallelogram condition is clearly equivalent to $AI = MD = \text{dist}(M, EF)$.
  • It's well-known that $AP = AQ = \frac{AB + AC}{2} = 12$.
  • Compute $AE = s - a = 5$.
  • Now, $$\cos^2 \left( \frac{A}{2} \right) = \frac{AN}{AI} = \frac{\text{dist}(A, EF)}{\text{dist}(M, EF)} = \frac{AE}{EP} = \frac{5}{7}$$where the penultimate equality follows from $EF \parallel \overline{MPQ}$.
  • Use $\triangle AEI$ to obtain $IF = \sqrt{10}$.
  • Extract $[ABC] = r \cdot s = 19 \sqrt{10}$, whence $m + n = \boxed{029}$.
This post has been edited 2 times. Last edited by ike.chen, Jul 25, 2023, 6:28 AM
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