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k a 2024 AMC 10/12B Math Jam!
LauraZed   0
Wednesday at 5:10 PM
Want to discuss the AMC 10/12B with AoPS instructors? Join us for the 2024 AMC 10B/12B Math Jam tonight, November 13, at at 7:30pm ET / 4:30pm PT, where we will discuss some of the most interesting problems from each test!

You can learn how to attend Math Jams here:
https://artofproblemsolving.com/school/mathjams

Finally, the Contests & Programs forum has also reopened for discussion of these exams, and you can find the official discussion threads here:
https://artofproblemsolving.com/community/c5h3442595_2024_amc_10b_discussion_thread
https://artofproblemsolving.com/community/c5h3442596_2024_amc_12b_discussion_thread
0 replies
LauraZed
Wednesday at 5:10 PM
0 replies
k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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0 replies
jlacosta
Nov 1, 2024
0 replies
A short one
jeffthebro   22
N 8 minutes ago by StarLex1
Let $x$ and $y$ be nonnegative real numbers that satisfy $x^2+y^2=4.$ Find the maximum value of
$$P=x^3+y^3.$$
22 replies
jeffthebro
Jun 12, 2022
StarLex1
8 minutes ago
Limit of sequence
belugacat   2
N 9 minutes ago by belugacat
For a positive integer $n$, let
$$u_n=n\left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right).$$Find all real numbers $a$ such that $\mathop {\lim }\limits_{n \to \infty } {n^a}{x_n}$ exist and different from 0.
2 replies
belugacat
Nov 8, 2024
belugacat
9 minutes ago
Inequalities
sqing   0
30 minutes ago
Source: Own
Let $  x\geq 0 .$ Prove that
$$ \frac{(\sqrt{9+4x}-\sqrt{ x })\sqrt{1+4x}}{9+4x}\leq  \sqrt{\frac{23}{128}+\frac{1}{8\sqrt{3}}}$$$$ \frac{(\sqrt{9+4x}-\sqrt{ x })\sqrt{3+8x}}{9+4x}\leq  \frac 1{4}\sqrt{\frac{33}{5}+2\sqrt{\frac{3}{5}}}$$
0 replies
sqing
30 minutes ago
0 replies
Reflections in a trapezoid
Shayan-TayefehIR   1
N 33 minutes ago by MathLuis
Source: IGO 2024 Intermediate Level - Problem 5
Point $P$ is the intersection of diagonals $AC,BD$ of the trapezoid $ABCD$ with $AB \parallel CD$. Reflections of the lines $AD$ and $BC$ into the internal angle bisectors of $\angle PDC$ and $\angle PCD$ intersects the circumcircles of $\bigtriangleup APD$ and $\bigtriangleup BPC$ at $D'$ and $C'$. Line $C'A$ intersects the circumcircle of $\bigtriangleup BPC$ again at $Y$ and $D'C$ intersects the circumcricle of $\bigtriangleup APD$ again at $X$. Prove that $P,X,Y$ are collinear.

Proposed by Iman Maghsoudi - Iran
1 reply
Shayan-TayefehIR
Yesterday at 7:35 PM
MathLuis
33 minutes ago
No more topics!
Pal triangles
v_Enhance   11
N 2 hours ago by Mathandski
Source: USA January TST for IMO 2013, Problem 1
Two incongruent triangles $ABC$ and $XYZ$ are called a pair of pals if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-element sets of pairwise relatively prime integers.

Determine if there are infinitely many pairs of triangles that are pals of each other.
11 replies
v_Enhance
Jul 30, 2013
Mathandski
2 hours ago
Pal triangles
G H J
Source: USA January TST for IMO 2013, Problem 1
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v_Enhance
6836 posts
#1 • 4 Y
Y by bobjoe123, HamstPan38825, Adventure10, Rounak_iitr
Two incongruent triangles $ABC$ and $XYZ$ are called a pair of pals if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-element sets of pairwise relatively prime integers.

Determine if there are infinitely many pairs of triangles that are pals of each other.
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polya78
105 posts
#2 • 5 Y
Y by langkhach11112, Aryan-23, Adventure10, Mango247, and 1 other user
I believe there are an infinite number of pals. If we complete the parallelogram, we see that the triangle with side lengths $AB, 2*AM, AC$ has twice the area of each triangle. So we are done if we can find and infinite number of relatively prime triples $(a,b,c)$ such that $a<b<c$ and triangles with side lengths $(a,2b,c),(2a,b,c)$ have the same area.

Using the area formula $\sqrt{s(s-x)(s-y)(s-z)}$ and crunching away, we wind up with the formula $2c^2=5(a^2+b^2)$. Now if $(n,m)$ satisfy $(2n+1)^2+1=2m^2$ (and there are an infinite number of such pairs), then it follows that we can let $a=2n+3,b=4n+1,c=5m$. Since $c$ is approximately $7n$, the relevant triangle inequalities are satisfied. The only case where where we don't have relatively prime numbers is when the common divisor is 5, and we can just forget about those cases or reduce by a factor of 5.
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Idio-logy
206 posts
#3
Y by
There are an infinite number of pals. Let the common three-element set be $\{a,b,c\}$. By Stewart's theorem, if we choose two of them (say $b$ and $c$) as two sides and $a$ be the median, the remaining side of the triangle has length $l=\sqrt{2(b^2+c^2-2a^2)}.$ So by the area formula
\begin{align*}
    \Delta &= \frac14\sqrt{-l^4-b^4-c^4+2l^2b^2+2b^2c^2+2c^2l^2}\\
    &= \frac14\sqrt{-4(b^2+c^2-2a^2)^2-b^4-c^4+4(b^2+c^2)(b^2+c^2-2a^2) + 2b^2c^2}\\
    &= \frac14\sqrt{-16a^4-b^4-c^4+8a^2b^2+8a^2c^2+2b^2c^2}.
\end{align*}Without loss of generality, two triangles have $\{a,b\}$ and $\{b,c\}$ as sides respectively. Then $$-16a^4-b^4-c^4+8a^2b^2+8a^2c^2+2b^2c^2 = -16c^4-b^4-a^4+8c^2b^2+8a^2c^2+2b^2a^2$$which is equivalent to $2b^2=5a^2+5c^2$.
Consider any solution to $(10u+3)^2-10v^2=-1$ (there are infinitely many by recursion), and take $(a,b,c) = (6u+1, 5v, 8u+3)$. It's not hard to verify that the resulting triangles satisfy the triangle inequality, and $a,b,c$ are pairwise coprime.
This post has been edited 1 time. Last edited by Idio-logy, Dec 27, 2020, 4:11 AM
Reason: fix
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naman12
1358 posts
#4
Y by
I'm posting this because memorizing random formulas from the AMC actually helps.

The answer is affirmitive. Note that at least one side of $\triangle ABC$ and $\triangle XYZ$ are equal - let $AB=XY=x,AC=y,XZ=z$ without loss of generality. Then, by 2, we get that $AM=z$ and $XW=y$. Thus, using the median formula, we get
\[2z=2AM=\sqrt{2(AB^2+AC^2)-BC^2}=\sqrt{2(x^2+y^2)-z^2}\implies BC=\sqrt{2x^2+2y^2-4z^2}\]By symmetry, we get that
\[YZ=\sqrt{2x^2+2z^2-4y^2}\]We have the following lemma:
Lemma. For a general triangle with side lengths $a,b,c$, we have the area of that triangle is also given by
\[\dfrac 12\sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}\]Proof. We just need to show that this is equivalent to Heron's formula. We get that by the difference of squares formula
\begin{align*}
    \dfrac 12\sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}&=\dfrac 14\sqrt{4a^2c^2-(a^2+c^2-b^2)^2}\\
    &=\dfrac 14\sqrt{(a^2+c^2+2ac-b^2)(a^2+c^2-2ac-b^2)}\\
    &=\dfrac 14\sqrt{((a+c)^2-b^2)((a-c)^2-b^2)}\\
    &=\dfrac 14\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}
\end{align*}which is Heron's formula. $\blacksquare$

Now, we get that
\[[ABC]=\dfrac 12\sqrt{x^2y^2-\left(\dfrac{x^2+y^2-(2x^2+2y^2-4z^2)}{2}\right)^2}=\dfrac 14\sqrt{4x^2y^2-\left(4z^2-x^2-y^2\right)^2}\]By symmetry, we get
\[[XYZ]=\dfrac 14\sqrt{4x^2z^2-\left(4y^2-x^2-z^2\right)^2}\]Equating, we get
\[4x^2(y^2-z^2)=4(x^2y^2-x^2z^2)=\left(4z^2-x^2-y^2\right)^2-\left(4y^2-x^2-z^2\right)^2=-5(3z^2+3y^2-2x^2)(y-z)(y+z)\]Assuming $y\neq z$, we get
\[-4x^2=15z^2+15y^2-10x^2\]which means that
\[2x^2=5y^2+5z^2\]Let $z=1$, and then we get
\[y^2-10\left(\dfrac x5\right)^2=1\]which has infinitely many solutions as a Pell equation. Thus, we can take any solution of
\[k^2-10\ell^2=1\]and use $(x,y,z)=(5\ell,k,1)$. We still need that $x^2+y^2\geq z^2$ and $x^2+z^2\geq y^2$. As $x$ and $y$ are positive integers, the first is obviously satisfied. For the second, we note that it is equivalent to
\[25\ell^2+1\geq k^2=1+10\ell^2\]which is definitely true. Thus, taking the pair mentioned above indeed is a solution, and as there are an infinitely many $(k,\ell)$, the problem statement is affirmitive.

Remark. I don't believe it! This information (the Lemma) was only memorized for the AMC! I'm actually surprised that it's on the TST - I'm glad for the AMC now (for the first time). Also, I pity those who don't know this formula.
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cyshine
236 posts
#5 • 1 Y
Y by PRMOisTheHardestExam
I gave this problem to my students last week, and I realized that the solutions here choose one of the elements of the set $\{AB,AC,AM\}=\{XY,XZ,XW\}$ equal to $1$, or state that since $c\approx 7n$ the triangle inequality is satisfied (which is not true because $a+b=6n+4<7n$.)
So here's a solution that works, without ever using Pell's equation. First, let me explain why setting an element equal to $1$ does not work. First, no triangle with integer sides can have one of them equal to $1$ unless it is isosceles (otherwise if $m>n$ are the other sides then $m\ge n+1$, violating the triangle inequality). Now, in most cases the median cannot be $1$: if you complete the parallelogram $ABDC$ (that is, define $D=B+C-A$), you'll see that you can construct a triangle with two sides $AB$, $AC$ and $A$-median $AM$ iff you can construct a triangle $\Delta$ with three sides $AB$, $AC$ and $2AM$ (both constructions are half of the parallelogram, and by the way, both triangles have the same area, so you can skip all the computations with medians.) If the median is $1$, then one side of $\Delta$ is $2$, and you either have an isosceles triangle with sides $2,a,a$ and area $\sqrt{a^2-1}$ or a triangle with sides $2,a,a+1$ and area $\frac1{16}\sqrt{3(4a^2+4a-3)}$, both of which are increasing with $a$.
For the sake of completion, let me rewrite the computations: let $AB=XY=x$, $AC=XW=y$ and $AM=XZ=z$ (otherwise the triangles are congruent). Then using the expanded form of Heron's formula $16S=2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4$ for $(a,b,c)=(x,y,2z)$ and $(x,2y,z)$ you can find
$2x^2y^2+8x^2z^2+8y^2z^2 - x^4 - y^4 - 16z^4 = 8x^2y^2 + 2x^2z^2 + 8y^2z^2 - x^4 - 16y^4 - z^4\iff 2x^2(z^2-y^2) = 5(z^2-y^2)(z^2+y^2)
\iff 2x^2 = 5y^2+5z^2.$
Since $5\mid x$, let $x=5t$, so $y^2+z^2=10t^2$. Then $y,z<\sqrt{10}t<5t=x$, and the largest side of each of the triangles with sides $x,y,2z$ or $x,2y,z$ is either $x$ or $2z$ in the first one, and either $x$ or $2y$ in the second one. We need then $x < y+2z$, $y+x > 2z$, $x<2y+z$, and $z+x > 2y$. That's why setting $y=1$ or $z=1$ is a bad idea: if $y=1$ then $z\approx \sqrt{10}t$ and $2z \approx 2\sqrt{10}t = \sqrt{40}t > 5t + 1 = x+y$, violating the triangle inequality.
What to do then? One neat trick you can do to try and keep $y$ and $z$ somewhat close to each other is taking advantage of the fact that $10=3^2+1^2$ and set $y=u+3v$ and $z=3u-v$ (we'll take care of signs later). Then
$y^2+z^2 = 10t^2\iff (u+3v)^2 + (3u-v)^2 = 10t^2 \iff u^2+v^2=t^2,$
and we can use Pythagorean triples! We can use the general form $(2mn; m^2-n^2; m^2+n^2)$, but to keep things simple we set $n=1$ and $m=2a$: $u=4a^2-1$, $v=4a$, and $t=4a^2+1$. Then $x=5t=5(4a^2+1)=20a^2+5$, $y=u+3v=4a^2+12a-1$, and $z=3u-v=12a^2-4a-3$. Then $x+y=24a^2+12a+4 = 2(12a^2+6a+2)>2z$, $x+z=32a^2-4a+2=2(16a^2-2a+1)>2z$, $2y+z=20a^2+20a-5>x$, and $y+2z=28a^2+4a-7>x$.
It remains to check whether $\gcd(x,y,z)=1$. Let $d=\gcd(4a^2+1,4a^2+12a-1)$. Then $d$ is odd and $d\mid (4a^2+12a-1-(4a^2-1))\iff d\mid 12a+2\iff d\mid 6a+1$. Then $d\mid 2a(6a+1)-3(4a^2+1)\iff d\mid 2a-3$, and $d\mid 6a+1-3(2a-3)\iff d\mid 10\iff d\mid 5$. We can just choose $a$ such that $4a^2+12a-1\not\equiv 0\pmod 5\iff -a^2+2a-1\not\equiv0\pmod 5\iff a\not\equiv 1\pmod 5$ (which also takes care of the factor $5$ in $x$.)
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anser
572 posts
#6
Y by
cyshine wrote:
I gave this problem to my students last week, and I realized that the solutions here choose one of the elements of the set $\{AB,AC,AM\}=\{XY,XZ,XW\}$ equal to $1$, or state that since $c\approx 7n$ the triangle inequality is satisfied (which is not true because $a+b=6n+4<7n$.)

If WLOG $AB = XY = c$, $AM = XZ = a$, and $AC = XW = b$, then we get $2c^2 = 5a^2+5b^2$. By the triangle inequality on $\triangle AMC$ and $\triangle AMB$, $a+b > \frac{1}{2}BC > |b-a|$ and $c+a > \frac{1}{2}BC > c-a$. So $a+b > \frac{1}{2}BC > c-a$ and $2a+b > c$. One can also verify that as long as $2a+b > c$, $\frac{1}{2}BC = \sqrt{\frac{1}{2}b^2 + \frac{1}{2}c^2 - a^2}$ satisfies the triangle inequalities (so it's a necessary and sufficient condition). Analogously in $\triangle XYZ$ we need $2b + a > c$. So I believe that we can't set $a$ or $b$ equal to 1, but the first solution is correct.
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Idio-logy
206 posts
#7
Y by
Quote:
First, let me explain why setting an element equal to $1$ does not work.
Indeed, my original solution was wrong. Triangle inequality implies that one would need $a+b > 2c > b-a$ and $b+c > 2a > b-c$ for the construction to work, but my original construction does not satisfy them. However, one could still use Pell equations (along the same lines as the first solution): consider any solution to $(10u+3)^2-10v^2=-1$ (there are infinitely many by recursion), and take $(a,b,c) = (6u+1, 5v, 8u+3)$. It's not hard to verify that they satisfy the size constraint and are pairwise coprime. Thanks for pointing out the mistake.
This post has been edited 1 time. Last edited by Idio-logy, Dec 27, 2020, 4:12 AM
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CyclicISLscelesTrapezoid
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Let $BC=a$, $CA=b$, $AB=c$, $YZ=x$, and $AM=m$. After analyzing config issues, we assume WLOG that $XY=m$, $XW=c$, and $XZ=b$. There's a way to derive $2b^2=5(c^2+m^2)$ without Heron's. Notice that since the areas of triangles $ABM$ and $XYW$ are the same, $\sin \angle BAM=\sin \angle YXW$. It's easy to check that $\angle BAM \neq \angle YXW$, so $\angle BAM+\angle YXW=180^\circ$. Then, we use LOC on $\triangle ABM$ and $\triangle YXW$ to get \[\frac{a^2}{4}=c^2+m^2-2cm \cos \angle BAC\]and \[\frac{x^2}{4}=c^2+m^2-2cm \cos \angle YXW=c^2+m^2+2cm \cos \angle BAC.\]Add these equations to get $a^2+x^2=8c^2+8m^2$. Then, plug in the median length formulas for $AM$ and $XW$ to prove that.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Dec 29, 2021, 5:19 AM
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IAmTheHazard
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The answer is yes, there do exist infinitely many pairs.

Suppose a triangle $PQR$ has sides $x,y$, and corresponding median with length $z$. Then the third side has length $\sqrt{2(x^2+y^2-2z^2)}$ by the midpoint formula. Then, expanding Heron's formula out and substituting, we find that
$$[PQR]=\frac{1}{4}\sqrt{-16z^4-x^4-y^4+8z^2x^2+8z^2y^2+2x^2y^2}.$$Now, WLOG let $AB=XY=a$, so $AC=XW=b$ and $AM=XZ=c$. The condition that $[ABC]=[XYZ]$ becomes
$$-16b^4-a^4-c^4+8b^2a^2+8b^2c^2+2a^2c^2=-16c^4-a^4-b^4+8c^2a^2+8b^2a^2+2a^2b^2,$$which simplifies to $2a^2=5b^2+5c^2$. Consider the equation $x^2-2y^2=-1$, over the nonnegative integers, which is a Pell equation that has infinitely many solutions, with minimal solution $(x,y)=(1,1)$. Evidently, any solution $(x,y)$ has $x$ odd, so we may take $(a,b,c)=(5y,2x-1,x+2)$ which we can verify satisfies the equation.
We now have to check that the triangle inequality holds. We can find that $BC=\sqrt{2((5y)^2+(2x-1)^2-2(x+2)^2)}$ and likewise $YZ=\sqrt{2((5y)^2+(x-1)^2-2(2x-1)^2)}$, and after some brute force computation we can show that we indeed get triangles with these values. Finally, we have to prove that infinitely many such $(a,b,c)$ are pairwise relatively prime. Throw out all $(x,y)$ such that $x \equiv 3 \pmod{5} \implies 5 \mid y$, which leaves an infinite number of $(x,y)$ still. Then $\gcd(2x-1,x+2)=\gcd(5,x+2)=1$. Further, we have $\gcd(x+2,5y)=\gcd(x+2,y)$, and if some prime $p$ divides both $x+2$ and $y$, then $x^2-2y^2 \equiv 4$, so $p \mid 5 \implies p=5$, contradiction. Likewise, $\gcd(2x-1,5y)=\gcd(2x-1,y)$, and if some prime $p$ divides them both then we have
$$4x^2-8y^2=-4 \implies 1 \equiv -4 \pmod{p} \implies p=5,$$which yields the same contradiction. Hence there are an infinite number of triples $(a,b,c)$ such that $\gcd(a,b)=\gcd(b,c)=\gcd(c,a)=1$, so we're done. $\blacksquare$

Remark: awful problem.
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HamstPan38825
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"You can't put computational problems on an olympiad..." —someone, probably

The answer is yes. Without loss of generality we may set $AB=XY=a$, $AM=XZ=b$, $AC=XN=c$. Now, the median formula gives the following two equalities:
\begin{align*}
2c^2+2a^2-BC^2 &= 4b^2 \\
2a^2+2b^2-YZ^2 &= 4c^2.
\end{align*}Now, equating both sides of the area of the two triangles yields $$a^2c^2 - \left(\frac{4b^2-a^2-c^2}2\right) = a^2b^2-\left(\frac{4c^2-a^2-b^2}2\right).$$This eventually simplifies to $$a^2 = \frac 52(b^2+c^2).$$Now, we may fix $c$ to get some solution to the Pell equation $$x^2-10y^2=1.$$It can be verified that these solutions also satisfy the triangle inequality.
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Martin2001
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Let $c<b<a,$ and WLOG let $AB=XY=a, AM=XZ=b,AC=XN=c.$ Note that $BC^2=2a^2+2c^2-4b^2,YZ^2=2a^2+2b^2-4c^2$ by Stewart's, so after a long computation with Heron we get that $2a^2=5(b^2+c^2).$ Now, let $(a,b,c)=(5m, 4n+1, 2n+3),$ where $(2n+1)^2+1=2m^2,$ which we have infinite solutions by Pell. Now, note that if we flip the triangle through $M,$ and let $A'=D,$ we have that triangle ABD has sidelengths either $a,2b,c$ or $a,b,2c,$ which both work by Triangle inequality as $a \approx 5\sqrt{2}n \approx=7n,$ and finally the only way for $a,b,c$ to not be coprime is when there is a common factor of $5,$ which we can either discard or divide out from all$.\blacksquare$
This post has been edited 1 time. Last edited by Martin2001, Jul 22, 2024, 5:48 PM
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Mathandski
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