A lot of integer lengths: JMO #6 or USAMO Problem 4

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BarbieRocks

1102 posts

BarbieRocks

#1 h Apr 29, 2010, 1:12 PM • 7 Y

Y by Davi-8191, OlympusHero, samrocksnature, HWenslawski, megarnie, Adventure10, Mango247

Let $ABC$ be a triangle with $\angle A = 90^{\circ}$ . Points $D$ and $E$ lie on sides $AC$ and $AB$ , respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$ . Segments $BD$ and $CE$ meet at $I$ . Determine whether or not it is possible for segments $AB$ , $AC$ , $BI$ , $ID$ , $CI$ , $IE$ to all have integer lengths.

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math154

4302 posts

math154

#2 h Apr 29, 2010, 1:20 PM • 4 Y

Y by samrocksnature, HWenslawski, Adventure10, Mango247

BarbieRocks wrote:

Then I made the sides $2rs, r^2-s^2, r^2+s^2$

It might not be primitive: this is basically what I did, but you have to do $2krs, k(r^2-s^2), k(r^2+s^2)$ .

Also, $1^2+7^2=2\cdot5^2$ .

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1=2

4594 posts

1=2

#3 h Apr 29, 2010, 1:21 PM • 4 Y

Y by Dmute, samrocksnature, Adventure10, Mango247

I had a really awesome trig solution:
(Proof by contradiction)
The angle bisectors of the triangle meet at the incenter of the triangle, so I is the incenter of the triangle. Draw perpendiculars from I to AC and AB. Call the feet F and G, respectively. If the inradius is $r$ , then $IF=IG=r$ . Let $\angle ABD=\alpha$ . Note that $\angle ACE=45-\alpha$ , because $\angle B+\angle C=90$ . We then have $IB=r*csc(\alpha)$ and $GB=r*cot(\alpha)$ . We are given that $BI$ and $ID$ are integers, so $\frac{BI}{ID}=\frac{BG}{GA}$ is rational, so $\frac{r*cot(\alpha)}{r}=cot(\alpha)$ is rational. We are given that $AB=r(1+cot(\alpha))$ is an integer, so $r$ is also rational. We're given that $BI=r*csc(\alpha)$ is rational, so $csc(\alpha)$ is rational. Therefore $\frac{1}{csc{\alpha}}=\sin{\alpha}$ and $\frac{cot(\alpha)}{csc(\alpha)}=\cos{\alpha}$ are rational. By similar reasoning using $\angle ACE$ , $\sin{45-\alpha}$ is rational. However, if we use the sine addition/subtraction formula, we get that $\frac{\sqrt{2}}{2}(\cos{\alpha}-\sin{\alpha})$ must be rational, which implies $\frac{\sqrt{2}}{2}$ is rational, which is clearly false. We have a contradiction, so at least one of those lengths must be non-integral.

@math154: You don't have to use $k$ unless you're assuming that $r$ and $s$ are relatively prime. Every right triangle has legs that can be expressed as $2rs$ and $r^2-s^2$ and a hypotenuse of $r^2+s^2$ for some $r$ and $s$ .

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math154

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math154

#4 h Apr 29, 2010, 1:23 PM • 29 Y

Y by djb86, MSTang, joey8189681, huricane, Delray, anser, mathisawesome2169, Relkuyh, ETS1331, rashah76, myh2910, Excursion, OlympusHero, samrocksnature, SPHS1234, megarnie, EpicBird08, Adventure10, Mango247, centslordm, H_Taken, and 8 other users

Eh, the best solution is Law of Cosines on $\angle BIC$ :
$BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,$ and we're done.

Edit:

1=2 wrote:

@math154: You don't have to use $k$ unless you're assuming that $r$ and $s$ are relatively prime. Every right triangle has legs that can be expressed as $2rs$ and $r^2-s^2$ and a hypotenuse of $r^2+s^2$ for some $r$ and $s$ .

Blargh no, read this.

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1=2

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1=2

#5 h Apr 29, 2010, 1:24 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Wow, that is a beastly solution.

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archimedes1

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archimedes1

#6 h Apr 29, 2010, 1:26 PM • 5 Y

Y by djb86, ayushk, samrocksnature, Adventure10, Mango247

Outline

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BarbieRocks

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BarbieRocks

#7 h Apr 29, 2010, 1:28 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

I got it wrong.

How would you continue with my solution?

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BarbieRocks

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BarbieRocks

#8 h Apr 29, 2010, 1:31 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

math154 wrote:

Eh, the best solution is Law of Cosines on $\angle BIC$ :
$BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,$ and we're done.

Edit:

1=2 wrote:

@math154: You don't have to use $k$ unless you're assuming that $r$ and $s$ are relatively prime. Every right triangle has legs that can be expressed as $2rs$ and $r^2-s^2$ and a hypotenuse of $r^2+s^2$ for some $r$ and $s$ .

Blargh no, read this.

Well, won't doing it for primitive pythagorean triples be enough, because you can dilate by a factor? Because you proved a side was irrational...

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math154

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math154

#9 h Apr 29, 2010, 1:35 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Well yeah, but it's easier to write up if you just do it generally... basically after the substitution just show that $CI/BI$ is a rational thing that I forgot times $\sqrt{2}$ .

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RminusQ

172 posts

RminusQ

#10 h Apr 29, 2010, 1:39 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

math154 wrote:

Eh, the best solution is Law of Cosines on $\angle BIC$ :
$BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,$ and we're done.

That is an awesome solution.

Quote:

Edit:

1=2 wrote:

@math154: You don't have to use $k$ unless you're assuming that $r$ and $s$ are relatively prime. Every right triangle has legs that can be expressed as $2rs$ and $r^2-s^2$ and a hypotenuse of $r^2+s^2$ for some $r$ and $s$ .

Blargh no, read this.

Crap, that's right. I just said assume they're all rational; we can express it as 2mn, m^2-n^2, m^2+n^2 for some rational m,n; but that might not be the case. Essentially, my proof was to show that:
$BI/ID = BC/CD = \frac{a}{\frac{ab}{a+c}} = \frac{a+c}{b}$ is rational; hence BC = a must be. From there, triangles ABD and ACE produce $b^2 + (a+c)^2$ and $c^2 + (a+b)^2$ as right triangles, with $a^2 = b^2+c^2$ . If you throw a k into the 2mn, m^2-n^2, m^2+n^2, it will still cancel, leaving you with both $2(m^2+n^2)$ and $m^2+n^2$ as squares of rational numbers.

i thought i remembered tex...

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kitsune

132 posts

kitsune

#11 h Apr 29, 2010, 1:45 PM • 4 Y

Y by Excursion, samrocksnature, Adventure10, Mango247

math154 wrote:

Eh, the best solution is Law of Cosines on $\angle BIC$ :
$BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,$ and we're done.

That's what I found after an hour, and I was like no way... this is too easy. I guess the hard part is actually finding that argument; I spent too much time on Pythagorean triples and angle bisectors.

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1=2

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1=2

#12 h Apr 29, 2010, 1:46 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Tip: Angle bisector theorem is useful only when you have information about AD, DC, AE, or EB.

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RminusQ

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RminusQ

#13 h Apr 29, 2010, 2:24 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

1=2 wrote:

Tip: Angle bisector theorem is useful only when you have information about AD, DC, AE, or EB.

Not necessarily. First off, you could use ABT to find those four lengths in terms of a,b,c. Second, you can use ABT on $\triangle DBC$ or $\triangle BEC$ .

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1=2

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1=2

#14 h Apr 29, 2010, 2:32 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Hmm.... true.

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blackbelt14253

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blackbelt14253

#15 h Apr 29, 2010, 2:33 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Hmm my solution was a 2.5 hour bash with ABT and Pythagoras.
So I basically labeled all the lengths, assumed all the said lengths were integers so that each "half" of each leg had to be rational, and so did the hypotenuse. Then it was a bunch of pythagorean triple bashing and basically the two angle bisectors can't both have rational lengths because $\sqrt{k}$ and $\sqrt{2k}$ cannot both be rational.

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Taco12

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Taco12

#73 h Jan 27, 2023, 9:19 PM

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Note that $\angle BIC=135^{\circ}$ . Thus, $\cos \angle BIC = -\frac{\sqrt2}{2}$ . LoC on $\triangle BIC$ now gives $BI^2+CI^2-BI\cdot CI \cdot\sqrt2 = AB^2+AC^2,$ a contradiction.

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samrocksnature

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samrocksnature

#74 h Jan 27, 2023, 9:33 PM

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Taco12 wrote:

Note that $\angle BIC=135^{\circ}$ . Thus, $\cos \angle BIC = -\frac{\sqrt2}{2}$ . LoC on $\triangle BIC$ now gives $BI^2+CI^2-BI\cdot CI \cdot\sqrt2 = AB^2+AC^2,$ a contradiction.

No bary?

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Taco12

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Taco12

#75 h Jan 27, 2023, 9:46 PM • 2 Y

Y by samrocksnature, Danielzh

samrocksnature wrote:

Taco12 wrote:

Note that $\angle BIC=135^{\circ}$ . Thus, $\cos \angle BIC = -\frac{\sqrt2}{2}$ . LoC on $\triangle BIC$ now gives $BI^2+CI^2-BI\cdot CI \cdot\sqrt2 = AB^2+AC^2,$ a contradiction.

No bary?

Why am I doing this...

Apply barycentric coordinates on $\triangle ABC$ . Note that $a=\sqrt{b^2+c^2}$ , so $I=(b^2+c^2:b^3+bc^2:b^2c+c^3)$ . Cevian parameterization stuff then gives $D=(b^2+c^2:0:b^2c+c^3), E=(b^2+c^2:b^3+bc^2:0)$ . Distance formula now yields a contradiction.

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Mathlover_1

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Mathlover_1

#76 h Apr 7, 2023, 6:59 AM • 1 Y

Y by samrocksnature

Can we solve this problem by cartesian coordinates?

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Infinity_Integral

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Infinity_Integral

#77 h Jun 14, 2023, 12:50 PM

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Mathlover_1 wrote:

Can we solve this problem by cartesian coordinates?

Using Cartesian Coordinates when the problem has a incentre and 2 non perpendicular angle bisectors and 4 lines involving these stuff is probably not a good idea, but the messier it gets the more likely it is to be irrational.

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trk08

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trk08

#78 h Jun 19, 2023, 9:47 PM

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We can say that $I$ must be the incenter of $\triangle{ABC}$ . This means that $AI$ bisects $\angle{BAC}$ , so $\angle{BAI}=45^{\circ}$ . If we use LoC on $\triangle{BAI}$ , we find that:
$AI^2+AB^2-2AB\cdot AI\cos{45}=BI^2.$
Suppose that all of these lengths are integers. As $\cos{45}=\frac{\sqrt2}{2}$ , $BI^2$ is irrational so $BI$ is not integer. This is a contradiction which means that not all of these side lengths can be integers.

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Infinity_Integral

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Infinity_Integral

#79 h Jul 28, 2023, 10:05 PM

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The cosine rule solution is really nice, but I just set all the lengths to be integer and length bash until I get sqrt2 is rational. This is a very nice Geom question.

Full proof here
https://infinityintegral.substack.com/p/usajmo-2010-contest-review

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huashiliao2020

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huashiliao2020

#80 h Jul 31, 2023, 11:46 PM

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just posting my scratch work with lpieleanu, oops i dont wanna do writeup but anyways the thing in diagram is sufficient to understand

Attachments:

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chakrabortyahan

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chakrabortyahan

#81 h Aug 12, 2023, 2:45 PM

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Let FSOC , if possible every given length is an integer . We use the fact that $AD= \sqrt{bc-mn}$ where $AD$ is the internal angle bisector of $\Delta ABC$ (with $D \in BC$ ) and $BD = m;CD=n$ and $b ,c$ are as usual length of the sides $AC$ and $AB$ .(This can be easily proved with help of the stuart's theorem . Then these are some of the required lengths :
$CE = \sqrt{ab-\frac{abc^2}{(a+b)^2}}$ $BD = \sqrt{ac-\frac{acb^2}{(a+c)^2}}$ $BI = \frac{a+c}{a+b+c} BD$ $ID = \frac{b}{a+b+c} BD$ and thus if both $BI$ and $ID$ are integers then so is $BD$ .So , $\frac{b}{a+b+c}$ has to be rational and so $(a+b+c)$ has to be rational and so $a$ has to be rational and as $a^2 = b^2+c^2$ so $a$ must be an integer.
Now by the property of pythagorean triplets we write $a , b , c$ in the form $g(r^2+s^2),2grs , g(r^2-s^2)$ where $r , s$ are coprime numbers with different parity .As , $CE$ is integer so $ab(a+b+c)(a+b-c)$ has to be a perfect square dividing the thing by $g^4$ will give us another perfect squarewriting in terms of $r,s$ we get $(r^2+s^2) 8r^2s^2(r+s)^2$ is perfect square and so $(r^2+s^2)2$ is a perfect square but as we have $r^2+s^2$ odd ,hence contradiction and as $CE$ is not an integer so at least one of $CI,IE$ must be a non-integer. $\blacksquare$

This post has been edited 6 times. Last edited by chakrabortyahan, Mar 21, 2024, 12:01 PM

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joshualiu315

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joshualiu315

#82 h Nov 19, 2023, 7:45 AM

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The answer is $\boxed{\text{no}}$ . Notice

$\angle BIC = \angle BAC+ \angle ACE + \angle ABD = 135^\circ.$
Hence,

$AB^2+AC^2=BC^2 = BI^2+CI^2+BI \cdot CI \cdot \sqrt{2},$
a contradiction as $\sqrt{2}$ is irrational.

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megahertz13

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megahertz13

#83 h Jan 31, 2024, 11:49 PM

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Video Solution in 3 minutes!

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megahertz13

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megahertz13

#84 h Nov 4, 2024, 3:12 AM

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The answer is no.

Note that $\angle{BIC}=90^\circ+\frac{\angle{A}}{2}=135^\circ.$ Law of Cosines on $\triangle{BIC}$ gives $BI^2+CI^2+\sqrt{2}\cdot BI\cdot CI=BC^2=AB^2+AC^2\implies \sqrt{2}=\frac{AB^2+AC^2-BI^2+CI^2}{BI\cdot CI}.$ If all of these segments have integer length, the left-hand side would be irrational, while the right-hand side is rational. Therefore, it is impossible for all of these segments to have integer length.

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gladIasked

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gladIasked

#85 h Nov 28, 2024, 9:38 PM

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dumb ahh solution:

The answer is no.

Assume for the sake of contradiction that there exists $\triangle ABC$ such that each of the given segments has integer length. First, note that $\angle IBC + \angle ICB=45^\circ$ , so $\angle BIC = \angle DIE = 135^\circ$ . Therefore, $\angle BIE = \angle CID = 45^\circ$ . Now, $\triangle DIC\sim \triangle IAC$ and $\triangle BIE\sim \triangle BAI$ by AA similarity. We are then able to derive the following equalities:
$\begin{align*} \frac{BE}{BI}=\frac{BI}{AB}=\frac{EI}{AI}\\ \frac{CD}{CI}=\frac{DI}{AI}=\frac{CI}{AC}. \end{align*}$ Thus, $BE=\frac{BI^2}{AB}$ , so $BE$ is rational. Analogously, $CD$ , $AE$ , and $AD$ are also rational. Also, $AI =\frac{EI\cdot BI}{BE}$ , so $AI$ is rational. By the Angle Bisector Theorem, $\frac{BC}{CA}= \frac{BE}{AE}\implies BC=CA\cdot \frac{BE}{AE}$ , so $BC$ is also rational. To finish, note that $[ABC] = \frac{bc}2$ , so by $A=rs$ we have the inradius $r=\frac{bc}{a+b+c}$ is rational. However, $AI = r\sqrt 2 = \frac{bc}{a+b+c}\cdot \sqrt 2$ , which is irrational, a contradiction. Therefore, no such triangle with the given conditions exists. $\blacksquare$

This post has been edited 2 times. Last edited by gladIasked, Nov 28, 2024, 9:45 PM

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Jupiterballs

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Jupiterballs

#86 h Dec 27, 2024, 12:00 PM

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We claim that the answer is no,
We prove our claim by contradiction,

Assume that $AB,AC,IC,IB$ $\in$ $\mathbb{Z}$

Now Let $\angle ABC$ = $2\theta$

Then, $\angle DBC$ = $\theta$

And as $\angle ACB$ = $90^\circ- 2 \theta$

$\angle ECB$ = $45^\circ- \theta$

So, $\angle BIC$ = $180^\circ$ - $\angle ACB$ - $\angle ECB$ = $135^\circ$

Now, by cosine law, we get that

$IB^2 + IC^2-\sqrt{2}\cdot IB\cdot IC$ = $AB^2 + AC^2$

Which implies that if all $AB,AC,IC,IB$ $\in$ $\mathbb{Z}$ , then $\sqrt{2} \in \mathbb{Z}$ , which is absurd.
$\mathbb{QED}$
$\blacksquare$

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Maximilian113

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Maximilian113

#87 h Mar 28, 2025, 10:25 PM

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Bro, I didn't see the Cosine Law and instead I did Sine Law howw :wallbash_red:

Let $x=AB, y=AC.$ It is well-known that $\angle BIC = 90^\circ + \frac12 \angle A = 135^\circ.$ Therefore by the Sine Law and Half Angle formula $IB = \sqrt{2} BC \sin \frac{\angle C}{2} = \sqrt{\sqrt{x^2+y^2}(\sqrt{x^2+y^2}-AC)}.$ Therefore if $\sqrt{x^2+y^2} \notin \mathbb Z,$ we have the square root of an integer minus an irrational, which clearly cannot be an integer.

Hence $AB, AC, BC$ are integers. Let $z=BC.$ Then for positive integers $m, n, k$ with $m > n,$ WLOG $x=2mn, y=m^2-n^2, z=m^2+n^2.$ Then $\sqrt{z^2-xz}, \sqrt{z^2-yz} \in \mathbb Z.$ The first one yields $\sqrt{(m^2+n^2)(m-n)^2} \in \mathbb Z \implies \sqrt{m^2+n^2} \in \mathbb Z,$ but the second one gives $\sqrt{(m^2+n^2)(2n^2)} \in \mathbb Z \implies \sqrt{2n^2} \in \mathbb Z,$ a contradiction. Hence the answer is no.

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