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

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 $s$.

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.

Wow, that is a beastly solution.

I got it wrong. How would you continue with my solution?

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...

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 $\sqrt{2}$.

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...

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.

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

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 $\triangle BEC$.

Hmm.... true.

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.

My too-bashy way: 1. 2mn, m^2+n^2, m^2-n^2 - find all the lengths from ABT and right triangles [because trig is for losers ] - everything cancels 2. Let L be the hypotenuse. Then L(L-B), L(L-C) and 2L are squares 3. From above, L must be square - contradiction

1. Using ABT and Pythagorean, BC must be an integer (if a is irrational, then so is BD, ehich implies that either BI or ID is irrational) Consider MNP, a triangle similar to ABC and a primitive Pythagorean triple. Then NQ and PR (analogous to BD and CE) are rational NQ has a length that is a rational number times $\sqrt{(m+n)^2+p^2}$; PR has a length that is a rational number times $\sqrt{(m+p)^2+n^2}$ m is odd, and so is exactly one of n and p, so one of $(m+n)^2+p^2$ and $(m+p)^2+n^2$ is 2 mod 4 and thus its square root (and corresponding length) is irrational, so one of those 4 segments touching the incenter has irrational length. Expand back out to ABC, and the one of those segments has irrational length.

right triangle ABD and the law of sines on AIC should do the trick

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. Woah holy cow I actually did LOC...now hoping for a 4.

BIC = BAC+ABD+ACE = 90+(ABD+ACE) = 90+(ABC+ACB)/2 = 135 so using LOC get BI^2+IC^2 + BI IC sqrt(2) = BC^2 if they are all integers then integer+integer+irrational=integer, which is false so they can't be integers

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.

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?

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.

Can we solve this problem by cartesian coordinates?

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.

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.

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

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

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. $\blacksquare$

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. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.794418642903586, xmax = 12.21505354692679, ymin = -1.1719411561946844, ymax = 9.147984881366593; /* image dimensions */ /* draw figures */ draw((0,0)--(6,0), linewidth(1)); draw((6,0)--(0,8), linewidth(1)); draw((0,8)--(0,0), linewidth(1)); draw((0,8)--(2.6666666666666665,0), linewidth(1)); draw((6,0)--(0,3), linewidth(1)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0.05455029479432724,0.12280971198643523), NE * labelscalefactor); dot((6,0),dotstyle); label("$B$", (6.045946469122638,0.12280971198643523), NE * labelscalefactor); dot((0,8),dotstyle); label("$C$", (0.05455029479432724,8.132494004361794), NE * labelscalefactor); dot((0,3),linewidth(4pt) + dotstyle); label("$D$", (0.05455029479432724,3.1058141631880347), NE * labelscalefactor); dot((2.6666666666666665,0),linewidth(4pt) + dotstyle); label("$E$", (2.7202138469319235,0.09742244006131523), NE * labelscalefactor); dot((2,2),linewidth(4pt) + dotstyle); label("$I$", (2.047451140916244,2.1030169221457946), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Video Solution in 3 minutes!

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.

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. $\blacksquare$

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. $\mathbb{QED}$ $\blacksquare$