Length bash to get $AD=\dfrac{abc}{b^2-c^2}.$ If $\gcd{(b,c)}=1$ then $b^2-c^2|a\implies a\ge (b+c)(b-c)>b+c,$ which is a contradiction. Thus $\gcd{(b,c)}>1$.

MathKnight16 wrote: Length bash to get $AD=\dfrac{abc}{b^2-c^2}.$ If $\gcd{(b,c)}=1$ then $b^2-c^2|a\implies a\ge (b+c)(b-c)>b+c,$ which is a contradiction. Thus $\gcd{(b,c)}>1$. Yep same method

Did something different. Let \(AD\ge 1\) wlog First I assumed \(gcd(AB, AC) = 1\)....(1) Then \(AD^2 =DB. DC=DC^2 - BC. DC \ge BC^2 - BC. DC > AB^2 +AC^2 - BC. DC\) . So \(BC. DC > AB^2 +AC^2 - AD^2 > 0\) . Adding \(- AD^2 \) to... (1) Then \(AB^2 +AC^2 - AD^2 > 0\) . But \(1-AD^2 <0\) Contradiction.

Mate007 wrote: Did something different. Let \(AD\ge 1\) wlog First I assumed \(gcd(AB, AC) < 1\)....(1) Then \(AD^2 =DB. DC=DC^2 - BC. DC \ge BC^2 - BC. DC > AB^2 +AC^2 - BC. DC\) . So \(BC. DC > AB^2 +AC^2 - AD^2 > 0\) . Adding \(- AD^2 \) to... (1) Then \(AB^2 +AC^2 - AD^2 > 0\) . But \(1-AD^2 <0\) Contradiction. your assumption is wrong as gcd is never less than 1

.... I changed it then it is correct or not

Easy to get $AD = \frac{abc}{b^2-c^2}$. If $\gcd(b,c) = 1$ then $\gcd(b^2-c^2,bc) =1$. Thus $b^2-c^2\mid a$ or \[a\geq b^2-c^2=(b+c)(b-c)\geq b+c > a.\] Thus contradiction and we are done.

i showed that $BD$ does not divide $AD^2$ which is contradiction(for gcd=1) as $AD^2=BD*DC$ so the solution follows

TheDarkPrince wrote: Easy to get $AD = \frac{abc}{b^2-c^2}$. If $\gcd(b,c) = 1$ then $\gcd(b^2-c^2,bc) =1$. Thus $b^2-c^2\mid a$ or \[a\geq b^2-c^2=(b+c)(b-c)\geq b+c > a.\] Thus contradiction and we are done. This may be a stupid question, but what if $b-c < 1$. Then shouldn't $(b+c)(b-c)\leq b+c $

Arhaan wrote: TheDarkPrince wrote: Easy to get $AD = \frac{abc}{b^2-c^2}$. If $\gcd(b,c) = 1$ then $\gcd(b^2-c^2,bc) =1$. Thus $b^2-c^2\mid a$ or \[a\geq b^2-c^2=(b+c)(b-c)\geq b+c > a.\] Thus contradiction and we are done. A stupid question. But what if $b-c < 1$. Then shouldn't $(b+c)(b-c)\leq b+c $ This is not stupid (But maybe you are). We are given that $b,c\in\Bbb N$, so $|b-c|\in\Bbb N$. So if your suggestion holds, then $b=c$ and the tangent does not meet $BC$, which contradicts the question.

Arhaan wrote: A stupid question. But what if $b-c < 1$. Then shouldn't $(b+c)(b-c)\leq b+c $ $AC>AB$ or $b>c$ or $b-c\geq 1$ (integers).

Well no one posted how to solve for $AD$ : First: Complex or bary. Second: Length chase - Let $AD = x$, we have from similarity, $\frac{c}{DB} = \frac{b}{x}$ and by PoP you get \[x^2 = \frac{xc}{b}\left(\frac{xc+ab}{b}\right)\implies x = \frac{abc}{b^2-c^2}.\]

TheDarkPrince wrote: Arhaan wrote: A stupid question. But what if $b-c < 1$. Then shouldn't $(b+c)(b-c)\leq b+c $ $AB>AC$ or $b>c$ or $b-c\geq 1$ (integers). $AC>AB$

TheDarkPrince wrote: Well no one posted how to solve for $AD$ : First: Complex or bary. Second: Length chase - Let $AD = x$, we have from similarity, $\frac{c}{DB} = \frac{b}{x}$ and by PoP you get \[x^2 = \frac{xc}{b}\left(\frac{xc+ab}{b}\right)\implies x = \frac{abc}{b^2-c^2}.\] One can also use the concept of $A$-Apollonius circle. Basically if $E$ and $F$ are the foot of internal and external angle bisectors of $\angle BAC$, then $$AD=DE=\frac{EF}{2}=\frac{1}{2}(BE+BF)=\frac{1}{2} \left(\frac{ac}{b+c}+\frac{ac}{b-c} \right)=\frac{abc}{b^2-c^2}$$

what will be the cut off for north bihar?

trash removed

After $p^2=q^2+aq$ how can you deduce that $q$ is also an integer?

Guys I have a doubt from kV all India students will be selected or something else

trash removed

We know that $\angle ADB = B-C$ and $\angle DBA = \pi - B$. By Sine Rule, we have $$AD = \frac{AB\sin (\pi-B)}{\sin (B-C)} = \frac{c\sin B}{\sin B\cos C - \cos B\sin C} = \frac{bc}{b\cos C-c\cos B}$$ or $$AD = \frac{bc}{b\cdot \frac{a^2+b^2-c^2}{2ab} - c \cdot \frac{a^2+c^2-b^2}{2ac}} = \frac{abc}{b^2-c^2}$$Thus $b^2 - c^2 \mid abc$. In particular $b+c \mid abc$. If $\gcd(b,c) = 1$ then $\gcd(b+c,c) = \gcd(b+c,b) = 1$, and this forces $b+c \mid a$, which is impossible as $a < b+c$. Therefore $\gcd(b,c) > 1$, as desired.

I used a whole different way to compute $AD$. I showed that if $AE$ and $AF$ are the internal and external angle bisectors respectively with $E$ and $F$ on $BC$, then $D$ is the midpoint of $EF$. And $AEF$ is a right angled triangle. So $AD = \frac{EF}{2}$. $EF$ is very easy to compute as we know $EB=\frac{ac}{b+c}$ and $FB=\frac{ac}{b-c}$. Proving the first part is mere angle chasing.

mathetillica wrote: Guys my solution to p1 goes like this: let $gcd(AB,AC)=1$ then $AB \nmid AC$ but $\frac{AC}{AB} = \frac{AD}{DB}$ which implies that $DB \nmid AD$ but $AD>DB$ so $DB \nmid AD^2$ but $AD^2=DB*DC$ hence a contradiction. i have explained everything in the solution including how i arrived at the ratios but didn't prove that $DB$ is an integer although it must be an integer of which i have a proof. How many marks would i be awarded? If this was an INMO you could have expected a zero (they're harsh at checking). But since this is an RMO, they should give you ten marks in even the worst case.

MathPassionForever wrote: I used a whole different way to compute $AD$. I showed that if $AE$ and $AF$ are the internal and external angle bisectors respectively with $E$ and $F$ on $BC$, then $D$ is the midpoint of $EF$. And $AEF$ is a right angled triangle. So $AD = \frac{EF}{2}$. $EF$ is very easy to compute as we know $EB=\frac{ac}{b+c}$ and $FB=\frac{ac}{b-c}$. Proving the first part is mere angle chasing. I have already posted it before. See post #15.

MathPassionForever wrote: If this was an INMO you could have expected a zero (they're harsh at checking). But since this is an RMO, they should give you ten marks in even the worst case. Where do you get all this information from? I am very interested.

biomathematics wrote: MathPassionForever wrote: If this was an INMO you could have expected a zero (they're harsh at checking). But since this is an RMO, they should give you ten marks in even the worst case. Where do you get all this information from? I am very interested. From my friend who qualified RMO but was disqualified in the INMO.

math_pi_rate wrote: MathPassionForever wrote: I used a whole different way to compute $AD$. I showed that if $AE$ and $AF$ are the internal and external angle bisectors respectively with $E$ and $F$ on $BC$, then $D$ is the midpoint of $EF$. And $AEF$ is a right angled triangle. So $AD = \frac{EF}{2}$. $EF$ is very easy to compute as we know $EB=\frac{ac}{b+c}$ and $FB=\frac{ac}{b-c}$. Proving the first part is mere angle chasing. I have already posted it before. See post #15. Sorry, didn't see ALL the solutions!

We use barycentric coordinates by setting $\Delta ABC$ as reference triangle.Then,let $$ D=(0:y:z)$$.Since $D$ lies on tangent to $A$ so $$ b^2z+c^2y=0 $$which gives $$ y:z=-b^2:c^2 $$which implies that $$ D=(0:-b^2:c^2) $$which gives $$ D=(0,\frac{-b^2}{c^2-b^2},\frac{c^2}{c^2-b^2}) $$. This gives us the displacement vector $AD=(1,\frac{b^2}{c^2-b^2},\frac{-c^2}{c^2-b^2})$.Hence $|AD|^2=\frac{a^2b^2c^2}{(b^2-c^2)^2}$ which implies $AD=\dfrac{abc}{b^2-c^2}.$ and rest is contradiction by triangle inequality. My first successful bary bash(though it was straightforward)of RMO.

MathKnight16 wrote: Length bash to get $AD=\dfrac{abc}{b^2-c^2}.$ If $\gcd{(b,c)}=1$ then $b^2-c^2|a\implies a\ge (b+c)(b-c)>b+c,$ which is a contradiction. Thus $\gcd{(b,c)}>1$. Yep!

MathKnight16 wrote: Length bash to get $AD=\dfrac{abc}{b^2-c^2}.$ If $\gcd{(b,c)}=1$ then $b^2-c^2|a\implies a\ge (b+c)(b-c)>b+c,$ which is a contradiction. Thus $\gcd{(b,c)}>1$. what does 'length bash' mean ? how to prove it?

Here's an equivalent version if you consider my comment there. Dutch BxMO TST 2018 p2 https://artofproblemsolving.com/community/q3h1887617p12867749

Given: $AD$ is an integer and sides $a,b,c$ are integers. So, we will express everything in terms of $a,b,c.$ After some angle chase, we apply sin rule in $\triangle DBA$ to get: $$DB=\frac{c\sin C}{\sin B\cos C -\sin C\cos B}=\frac{c^2}{b\cos C - c\cos B}\underbrace{\implies}_\text{using cosine rule}\boxed{DB=\frac{ac^2}{b^2-c^2}}$$Now as $$DA^2=DB \times DC \implies DA^2=\left(\frac{ac^2}{b^2-c^2}\right) \left(a+\frac{ac^2}{b^2-c^2}\right) \implies DA = \frac{abc}{b^2-c^2}$$$$\boxed{AD=\frac{abc}{(b+c)(b-c)}}$$Suppose gcd$(b, c) = 1$. Then $\text{gcd}(b, b^2- c^2) = 1 = \text{gcd}(c, b^2-c^2)$. Now, since $AD$ is an integer $\implies b^2-c^2$ divides $a\implies b + c$ divides $a \implies a \geq b + c$. But we know that $b+c > a $ (by triangle inequality) So, gcd$(b, c) > 1$. $\blacksquare$

mathetillica wrote: Mate007 wrote: Did something different. Let \(AD\ge 1\) wlog First I assumed \(gcd(AB, AC) < 1\)....(1) Then \(AD^2 =DB. DC=DC^2 - BC. DC \ge BC^2 - BC. DC > AB^2 +AC^2 - BC. DC\) . So \(BC. DC > AB^2 +AC^2 - AD^2 > 0\) . Adding \(- AD^2 \) to... (1) Then \(AB^2 +AC^2 - AD^2 > 0\) . But \(1-AD^2 <0\) Contradiction. your assumption is wrong as gcd is never less than 1 It's okay.. he just assumed that $AB$ and $AC$ are coprime. Then went on to prove that his assumption leads to a contradiction.. Therefore, $gcd(AB,AC)$ is always greater than $1$. Seems fine to me.. where's the problem?

Alternate segment theorem and then done!!

so we denote $AB=c,AC=b$ and $BC=a$ also we denote $AD=x$ and $DB=k$ by pop we get $k(k+a)=x^2$ and by base division thereom we have $k=\frac{xc}{ab}$ so we plug in $k(k+a)=x^2$ to get $x=\frac{abc}{b^2-c^2}$ now since $x \in \mathbb{Z}$ we get $b^2-c^2|abc$ now FTSOC assume $\gcd(b,c)=1 \implies b^2-c^2|a$ but from triangle inequality we have $b+c>a$, hence we must have $b^2-c^2 > a$ but since $b^2-c^2|a$ this gives $b^2-c^2 \leqslant a$ we get a contradiction and hence $\gcd(AB,AC)>1$ $\blacksquare$

Use the Eulerian notation for the sides of $\triangle ABC$. Claim 1: $BD=\frac{abc}{b^2-c^2}$ Proof: We angle chase to get $\angle ADC = \angle ABC - \angle ACB$ We see that $\frac{AD}{sinB}=\frac{c}{sin(B-C)}$ Using law of sines and law of cosines to simplify this further, we can get that $AD=\frac{abc}{b^2-c^2}$. So claim $1$ is proved. Now $FTSOC$, assume that $\gcd(b,c)=1$. Using Euclidean reduction, we can get that $b+c,b-c$ do not divide $bc$. This gives that $b^2-c^2|a$. But by the triangle inequality, we must have $b+c>a$ and also we are given $b>c$. Thus this is not possible and $\gcd(b,c)>1$. Our proof is thus complete.