Sreejato's triangle

From the angle-bisector theorem, we know that $\frac{BD}{DC}=\frac{AB}{AC}=\frac{\sqrt{41}}{\sqrt{164}}=\frac{1}{2}.$ Thus we let $BD=x, DC=2x.$ Additionally, we can let $AD=y.$ From a direct application of Stewart's Theorem on $\triangle ABC$ , we obtain $82x+164x=3x(y^2+2x^2).$ That is, $y^2+2x^2=82.$ Since $y$ and $2x$ are both integers, we find that $(x,y)=(8,3)$ is the only solution which yields $AD+DC=y+2x=14.$

Let $\angle BAD = \angle DAC = \alpha, AD = x$ and $BD = y$

$\frac {BD}{DC} = \frac {area of BAD}{area of DAC} = \frac {\frac {1}{2} (AB)(AD) sin \alpha}{\frac {1}{2} (AC)(AD) sin \alpha} = \frac {\sqrt{41}}{\sqrt{164}} = \frac {1}{2}$

Using cosine rule, we have

$y^2 = 41 + x^2 - 2\sqrt{41}x cos \alpha - (1)$

$(2y)^2 = 164 + x^2 - 4\sqrt{41}x cos \alpha - (2)$

Multiplying (1) by 4 we get:

$4y^2 = 164 + 4x^2 - 8\sqrt{41}x cos \alpha - (3)$

Subtracting (2) from (3) we get:

$0 = 3x^2 - 4\sqrt{41}x cos \alpha$

$x = \frac{4}{3} \sqrt {41} cos \alpha$ (since $x$ is not zero)

$cos \alpha = \frac {3}{4\sqrt{41}}x - (4)$

Substituting (4) back into 2 we get:

$(2y)^2 = 164 + x^2 - 4\sqrt{41}x \frac {3}{4\sqrt{41}}x = 164 + x^2 - 3x^2 = 164 - 2x^2$

$2x^2 + (2y)^2 = 164$

Note that $AD = x$ and $DC = 2y$ .Trying $2y$ from integers 1 to 12 and finding the respective $x$ , we notice that only $2y = 6$ will produce an integer $x$ , which is $x = 8$ .

Therefore, $AD + DC = 8+6 = 14$

Interesting note: We can apply Stewart's theorem to this problem after realising that $\frac {BD}{DC} = \frac {1}{2}$ . Using the theorem, we can simplify the equation $AB^2 \times DC + AC^2 \times BD = BC \times (AD^2 + BD \times DC)$ to get $x^2 + 2y^2 = 82$ , which when multiplied by 2, gives us $2x^2 + 4y^2 = 164$ , similar to the equation above.

Let $DC = p$, $DC = q$, and $BC = a-p$. We first show that $a = \frac{3}{2} p$. Given that $\angle BAD = \angle CAD = \alpha/2$, then by the Law of Sines, $\frac{\sin \alpha/2}{p} = \frac{\sin \gamma}{q}$, where $\gamma = \angle ACB$. But we also have $\frac{\sin \gamma}{\sqrt{41}} = \frac{\sin \alpha}{a}$ for the whole $\triangle ABC$, so it follows that $\frac{\sin \alpha/2}{p} = \frac{\sqrt{41} \sin \alpha}{aq}$. Similarly by considering $\beta = \angle ABC$, we have $\frac{\sin \alpha/2}{a-p} = \frac{\sqrt{164} \sin\alpha}{aq}$. Now dividing these two results gives $\frac{a-p}{p} = \frac{\sqrt{41}}{\sqrt{164}} = \frac{1}{2}$, which shows our claim that $a = \frac{3}{2} p$, and in particular, $BD = p/2$.

Next we use the Law of Cosines to obtain $\frac{164 + q^2 - p^2}{2q\sqrt{164}} = \cos \frac{\alpha}{2} = \frac{41 + q^2 - (p/2)^2}{2q \sqrt{41}}$. This simplifies to the Diophantine equation $2q^2 + p^2 = 164$ for positive integers $p, q$. By the triangle inequality, we must have $6 = \sqrt{36} < \sqrt{41} < q < \sqrt{164} < \sqrt{169} = 13$, or $7 \le q \le 12$. Similarly, $6 < \sqrt{164} - \sqrt{41} < \frac{3}{2}p < \sqrt{41} + \sqrt{164} = 3 \sqrt{41}$, or $5 \le p \le 12$. Since our Diophantine condition requires $p$ be even (otherwise the LHS is odd), we need only check the possibilities $p \in { 6, 8, 10, 12 }$. We find the only solution is $p = 6$, $q = 8$, and therefore $AD + DC = p+q = 14$.

$AB = \sqrt{41}$ and $AC = \sqrt{164} = 2\sqrt{41}$
By law of internal angle bisector, we get $\frac{BD}{DC} = \frac{AB}{AC} = \frac{\sqrt{41}}{2\sqrt{41}} = \frac{1}{2}$
Therefore, $DC = 2BD$
Let $DC = x$ where $x$ is a positive integer
$BD = \frac{DC}{2} = \frac{x}{2}$
$BC = BD+DC = \frac{x}{2} + x = \frac{3}{2}x$
According to Stewart law, $AD^2 \times BC = AB^2 \times DC + AC^2 \times BD - BD \times DC \times BC$
$AD^2 \times \frac{3}{2}x = 41 \times x + 82 \times x - \frac{3}{4}x^3$
$AD^2 = 82-\frac{x^2}{2}$
Since $AD$ is a positive integer, then $x$ must be an even number or $x=2k$ where $k$ is a positive integer
$AD^2 = 82 - \frac{(2k)^2}{2} = 82-2k^2 = 2(41-k^2)$
Since $AD^2 = 2(41-k^2)$ , then $41-k^2 > 0$ or $k$ is a positive integer less than $7$
By trying $k=1,2, \dots,6$ , we get the only value of $k$ is $3$ such that $82-2k^2$ is a perfect square
$BC = \frac{3}{2}x = 3k = 9$
We can easily check that $\sqrt{41}, 2\sqrt{41}, 9$ are three sides of a triangle or there exists a triangle with sides $\sqrt{41}, 2\sqrt{41}, 9$
Therefore, $k=3$ satisfies the condition
$AD = \sqrt{82-2(3)^2} = 8$
$DC=x=2k=2(3)=6$
$AD+DC = 8+6 = 14$
And we are done

Suppose AD produced meets the circumcircle of ΔABC at P. Note that angle ABD= angle APC being angles in the same segment determined by AC. Since AD is the internal bisector of angle BAC, angle BAD= angle DAC. Thus we can conclude that triangles ABD and APC are similar. Thus, AB/AP=AD/AC. This implies AB.AC=AD.AP. Now AD^2=AD(AP-DP)= AD.AP-AD.DP. So AD.AP=AD^2+ AD.DP. Putting this value in the previously found equation, we obtain AB.AC=AD^2+ AD.DP. But from the Secant theorem, AD.DP=BD.DC. Putting this value, AB.AC=AD^2+ BD.DC. Note that from Internal Angle Bisector Theorem, AB/AC=BD/DC, which implies BD=(AB.DC)/AC . Plugging this value and multiplying both sides by AC/AB gives AC^2=(AD^2.AC)/AB+ DC^2. Let AD=x, DC=y. Substituting the values gives 164=2x^2+ y^2. It can easily be verified that the only positive integral solution to the equation is (x,y)=(8,6). This gives AD=8, DC=6. So, AD+DC=14.

When I first saw this problem I definitely thought it would utilize exterior right triangles and Fermat's theorem that $p = a^2 + b^2$ iff $p = 4k + 1$ . Oh well. Anyways, on to the solution.

Draw a triangle $ABC$ as indicated in the question. Because of the angle bisector theorem, $BD = x$ and $DC = 2x$ . Say $m\angle ADC = \theta$ and $m\angle ADB = 180 - \theta$ . Finally, say $AD = y$ . Next, use law of cosines:

$164 = y^2 + 4x^2 - 2(y)(2x)cos\theta$

$41 = y^2 + x^2 - 2(y)(x)cos(180 - \theta)$

Multiply the second equation by 2 and apply the identity $cos(180 - \theta) = -cos\theta$ to get $82 = 2y^2 + 2x^2 + 4xycos\theta$ Adding this equation and the first equation together we get $246 = 3y^2 + 6x^2 \rightarrow 82 = y^2 + 2x^2$ Since $y$ and $2x$ must be integers, then $x$ must also be an integer. The only solution that satisfies this equation is $(3, 8)$ . Thus, $x + 2y = 8 + 6 = 14$ .

For any triangle $ABC$ , we denote that $AB = c, BC = a, AC = b$ , $S_{ABC}$ is the square of triangle ABC and $l_{a}$ is the length of internal angle bisector of $ABC$ , then

$l_{a} = \dfrac {\sqrt{bc(b+c+a)(b+c-a)}}{b+c}$ .

There are a number of ways to prove this equation. Here is one of them.

$S_{ABC}=S_{ABD}+S_{ACD}$

$\Rightarrow bcsinA=c.ADsin\dfrac{A}{2}+b.ADsin\dfrac{A}{2}$

$\Rightarrow AD=\dfrac{2bc}{b+c}.cos\dfrac{A}{2}$

$\Rightarrow AD^2=\dfrac{2b^2c^2}{(b+c)^2}.(cosA+1)=\dfrac{2b^2c^2}{(b+c)^2}.\dfrac{(b+c)^2-a^2}{2bc}$

$\Rightarrow AD=\dfrac {\sqrt{bc(b+c+a)(b+c-a)}}{b+c}$ .

With $b=\sqrt{41}$ and $c=\sqrt{164}$ we have $l_{a}=\dfrac{\sqrt{2(369-a^2)}}{3}$ . This implies that the value of $l_{a}$ is an integer in $[1, 9]$ .

Because AD is the internal angle bisector of $ABC$ , we have $\dfrac{CD}{BC}=\dfrac{AC}{AB+AC}=\dfrac{2}{3}$ . In other side, the length of CD is an positive integer therefore the length of BC is a rational number.

Combining two results above, we can compute directly and see that there is only a value $l_{a}=8$ corresponding to a rational value of $a (a=9)$ . Therefore, $AD=8$ and it is easy to see that $CD=6$ .

Finally $AD+CD=14$ .

For simplicity, put $\sqrt{41}=a, BC=b$ . We have $\frac{DB}{DC} = \frac{AB}{AC} = 2$ . Then $DC = \frac{2}{3} b$ . We have $AD=\frac{2 \cdot AB \cdot AC \cdot \cos \frac{A}{2}}{AB+AC} = \frac{4}{3} \cdot a \cdot \cos \frac{A}{2}$ . On the other hand, we have $\cos\frac{A}{2} = \sqrt{\frac{(AB+AC+BC)(AB+AC-BC)}{2 \cdot AB \cdot AC}} = \sqrt{\frac{9a^2-b^2}{8a^2}}$ . Therefore, $AD = \frac{4}{3} \cdot a \cdot \sqrt{\frac{9a^2-b^2}{8a^2}} = \frac{1}{3} \cdot \sqrt{2(369-b^2)}$ . Thus, $AD+DC = \frac{1}{3} \left( \sqrt{2(369-b^2)} +2b \right)$ . Since the length of $CD$ is a positive integer, we deduce that $3 \mid b$ . Otherwise, by triangle inequality, we have $\sqrt{41} < b < 3\sqrt{41} \Rightarrow b \in \{ 9, 12, 15, 18\}$ . By calculating directly, we obtain that $b=3$ and $AD+DC=14$ .

Let lengths AD = d, BD = m, CD = n By Angle-Bisector thm, n=2m Then, applying Stewart's thm and simplifying, we get n^2+2d^2 = 164 Letting n=2k, d=2r, we get k^2+2r^2 = 41 k has to be odd. Testing for 1 and 3, we find (k,r) = (3,4) works. Therefore, (d,n) = (6,8), total value is 14.

We know that $\dfrac{DC}{BD}=\dfrac{AC}{AB}=\dfrac{\sqrt{164}}{\sqrt{41}}=2$ Let $BD=z$ , implies $DC=2z$ , and $AD=x$ . Then $z^2=41+x^2-2x\sqrt{41}\cos(\angle BAD)...(1)$ and $4z^2=164+x^2-2x\sqrt{164}\cos(\angle CAD)..(2)$ Since $\angle BAD = \angle CAD$ , Then eliminating $(2)$ with $2*(1)$ , we will get that $2z^2=82-x^2$ $x$ must be even, and $2\le x\le 8$ . For $x=2,4,6$ , we will get that $z$ is not integer. But for $x=8$ , we will get that $z=3$ . So $AD+DC=x+2z=8+6=14$

$\text{Angle Bisector Theorem gives,}\\ \dfrac{BD}{DC}= \dfrac b c= \dfrac{ \sqrt{41} }{ \sqrt{164} }= \dfrac 1 2. \\$ \therefore ~DC=2*BD. \implies \color{#3D99F6}{DC=\dfrac 2 3 *(BD+DC)=\dfrac 2 3 *a. ~~\implies ~~3|a}\\ AD=l_a=\dfrac{ \sqrt{bc*(b+c+a)(b+c _ a) } {b+c}\\ =\dfrac{ \sqrt{ \sqrt{41}*\sqrt{164}*(3*\sqrt{41} - a^2} } {3*\sqrt{41}.\\ =\dfrac{ \sqrt{ 2*(3*\sqrt{41} - a^2} } {3} \\ Since ~ l_a ~ is ~an ~ integer, ~ \sqrt{ 2*2*\dfrac{ (369 -a^2) }{18} } \text{must be a perfect square.}\\ For ~ this, ~ (369 -a^2) ~ must ~ be ~ even, ~ \implies ~a^2 ~ must ~ be ~ odd, ~ but ~ 3|a, ~ \therefore ~ a=3,9,…\\ For ~ a=3 ~ \dfrac{ (369 -a^2) }{18} ~ is ~ not ~ a ~ perfect ~ square.\\ For ~ a=9 ~ \dfrac{ (369 -a^2) }{18}=16.\\ \therefore ~ AD=l_a=\sqrt{2*2*16}=8. ~~~~~ DC=\dfrac 2 3 *a=6.\\ \implies ~ AD + DC=8 + 6 =\Large ~~~~\color{#D61F06}{14}

Obviously, $\frac{BD}{DC}=\frac{1}{2}$ , so let $BD =m \Rightarrow DC = 2m$ and $AD = k$ . Stewart's Theorem said that on $\Delta ABC$ , we have $82m+164m=3m(k^2+2m^2)$ $246m = 3m(k^2+2m^2$ $k^2+2m^2=82$ $k^2 = 82-2m^2$

In equation $k^2 = 82-2m^2$ , we conclude that $2|k$ . Let $k=2p$ , then we have $4p^2=82-2m^2$ $2p^2=41-m^2$ $m^2=41-2p^2$

We know that $41-2p^2\geq 0$ so $p\leq 4$ since $k$ is positive, so $p$ is positive too. Because $k, p$ are integers and $m$ is the positive multiple of $0.5$ (look at the relation $BD$ and $DC$ ), we may attempt trials and errors. For several attempts, we may have that for $p= 4 \Rightarrow m= 3$ and $k = 8$ . We know that $BD=m=3$ , so $DC = 2m = 6$ . Finally, $AD+DC = 8+6 =14$

take AD=y and DC=2x gives BD=x. Now apply cosine rule for angle ADC and ADB. Add them to get 2x^2+y^2=82. Which gives AD=8 and CD=6