Finding the Side Length

Firstly, draw $CD$ and $CE$ which trisect angle $BCA$ , and $D$ and $E$ are on line $AB$ with $D$ nearer to $A$ . Let $\angle CAB = \alpha$ , so $\angle BCE=\angle ECD = \angle DCA = \alpha$ . Then, we can get triangle $CEB$ is similar to triangle $ACB$ since they are $\alpha - 3\alpha - 180^\circ - 4 \alpha$ triangles. Also, triangle $DCE$ is similar to triangle $CAE$ , since they are $\alpha - 2 \alpha - 180^\circ - 3 \alpha$ triangles.

Since $CBD$ is an isosceles triangle, $CB=BD=343$ and $AD=DC=504-343=161$ . Since triangle $CEB$ is similar to triangle $ACB$ , we get $BE : BC = BC : BA$ which gives $BE= \frac {16807}{72}$ and $DE=343-BE= \frac {7889}{72}$ . Since triangle $DCE$ is similar to triangle $CAE$ , we get $DE:CE = CE:AE$ which gives $CE= \frac {12397}{72}$ . Also, $CD:AC = DE:CE$ which gives $161:AC= \frac {7889}{72} : \frac {12397}{72}$ . Solving it we get $AC = 253$ and we are done.

Let $\alpha = \angle CAB$ . Since $0 < \alpha < 180^\circ$ , we have $0 < \sin \alpha \leq 1$ . By Sine Rule, we get that $\frac {\sin \alpha} {343} = \frac {\sin (3 \alpha)} {504} = \frac {\sin(180-4\alpha)}{AB}$ .

By the triple-angle formula, we know that $\sin (3\alpha) = 3 \sin \alpha - 4 \sin^3 \alpha$ . Substitution into the first equality yields $\frac {\sin \alpha} {343} = \frac {3 \sin \alpha - 4 \sin^3 \alpha} {504}$ , which simplifies to $196 \sin^3 \alpha - 75\sin\alpha = 0$ , so $\sin \alpha = \frac {5 \sqrt{3}} {14}$ , and $\cos \alpha = \frac {11}{14}$ . (The other 2 roots $0$ and $-\frac {5 \sqrt{3} } {14}$ are rejected because of our condition on $\sin \alpha$ .

By the double-angle formula, we know that $\sin (180^\circ - 4 \alpha) = \sin (4 \alpha) = 2 \sin (2\alpha) \cos (2\alpha) = 4 \sin \alpha \cos \alpha ( 1 - 2 \sin ^2 \alpha )$ . Substituting into the second equality yields $\frac {\sin \alpha} {343} = \frac {4 \sin \alpha \cos \alpha (1-2 \sin^2 \alpha)} { AB}$ , which simplifies to $AB = 343\times 4 \cos \alpha (1-2\sin^2 \alpha)$ . Substituting the values for $\sin \alpha$ and $\cos \alpha$ above will give $AB = 253$ .

Let $\angle CAB=\alpha$ , so that $\angle BCA=3\alpha$ . Construct the point $D$ on $\overline{AB}$ such that $\angle ACD=\alpha$ . Note that this creates the isosceles triangle $ACD$ . We also have $\angle CDB=\angle DAC+\angle ACD=\alpha+\alpha=2\alpha=\angle BCD$ , so triangle $CDB$ is isosceles as well. This means $BD=BC=343$ , and so $CD=AD=AB-BD=504-343=161$ .

Now by dropping the perpendicular of $D$ to $\overline{AC}$ we see that $AC=2(161\cos \alpha)=322\cos \alpha$ . Hence we are done if we can find $\cos \alpha$ .

We consider triangle $CDB$ since we know all its side lengths and angles. By the law of sines, $\frac{BC}{\sin \angle CDB}=\frac{CD}{\sin \angle DBC}\implies \frac{343}{\sin 2\alpha}=\frac{161}{\sin(180^{\circ}-4\alpha)}$ Using trig identities, $\sin(180^{\circ}-4\alpha)=\sin 4\alpha=2\sin 2\alpha\cos 2\alpha$ , so after some rearranging the equation becomes $\cos 2\alpha=\frac{23}{98}$ . By the half angle formula for cosine, we have $\cos \alpha=\pm \sqrt{\frac{\cos 2\alpha+1}{2}}$ . But since $\alpha$ and $3\alpha$ are angles of triangle $ABC$ , we know $\alpha+3\alpha<180^{\circ}$ , or $\alpha<45^{\circ}$ . Therefore, $\cos \alpha$ is positive and $\cos \alpha=\sqrt{\frac{\cos 2\alpha+1}{2}}=\frac{11}{14}$ . The length we seek is $AC=322\cos \alpha=\boxed{253}$ .

DC=CE=sqrt(EB ED)=sqrt(343 (504+343))=539, DA=DB DE/DC=504 (504+343)/539=792, AC=AD-DC=792-539=253

Let $\angle CAB = \alpha$ . Then $\angle BCA = 3\alpha$ and $\angle ABC = 180 - 4\alpha$ .

By sine rule, we have $\frac{343}{sin \alpha}=\frac{504}{sin 3\alpha}=\frac{AC}{sin (180 - 4\alpha)}=\frac{AC}{sin 4\alpha}, sin \alpha \neq 0$ .

From $\frac{343}{sin \alpha}=\frac{504}{sin 3\alpha}$ , we have $\frac{sin 3\alpha}{sin \alpha}=\frac{504}{343} - (1)$ .

Then, note that

$sin 3\alpha = sin (2\alpha + \alpha)$

$= sin 2\alpha cos \alpha + cos 2\alpha sin \alpha$

$= 2 sin \alpha cos^2 \alpha + sin \alpha (1 - 2sin^2 \alpha)$

$= 2 sin \alpha (1 - sin^2 \alpha) + sin \alpha (1 - 2 sin^2 \alpha)$

$= 2 sin \alpha - 2 sin^3 \alpha + sin \alpha - 2 sin^3 \alpha$

$= 3 sin \alpha - 4 sin^3 \alpha$

Combining it with $(1)$ , we obtain $3 - 4 sin^2 \alpha = \frac{504}{343}\ \Rightarrow sin^2 \alpha = \frac{75}{196}$

From $\frac{343}{sin \alpha}=\frac{AC}{sin 4\alpha}$ , we have $\frac{sin 4\alpha}{sin \alpha}=\frac{AC}{343} \Rightarrow AC = 343 \times \frac{sin 4\alpha}{sin \alpha}$ .

Computing $\frac{sin 4\alpha}{sin \alpha}$ , we get

$\frac{sin 4\alpha}{sin \alpha} = \frac{2 sin 2\alpha cos 2\alpha}{sin \alpha}$

$= \frac{4 sin \alpha cos \alpha cos 2\alpha}{sin \alpha}$

$= 4 cos \alpha cos 2\alpha$

$= 4 (\sqrt {1 - sin^2 \alpha})(1 - 2 sin^2 \alpha)$

(since $0^\circ < \angle BAC + \angle BCA = 4\alpha < 180^\circ \Rightarrow 0^\circ < \alpha < 45^\circ$ )

Thus,

$AC = 343 \times 4 \times (\sqrt {1 - sin^2 \alpha}) (1 - 2 sin^2 \alpha)$

$= 1372 (\sqrt{1 - \frac{75}{196}}) (1 - 2(\frac{75}{196}))$

$= 1372 (\sqrt{\frac{121}{196}}) (\frac{23}{98})$

$=1372 (\frac{11}{14}) (\frac{23}{98})$

$= 253$

By Law of Sines we have sin(3x)/sinx=504/343=72/49. Expanding sin(3x) as 3sin(x)cos^2(x)-sin^3(x), we can substitute cos^2(x)=1-sin^2(x) and get sin(x)=5\sqrt{3}/14. Now since <B=180-4x, sinB=sin4x, so by Law of Sines again, we have AC/343=sin(4x)/sinx.

Since sin(4x)=2sinxcosxcos(2x), we have 2 343 cos(x)cos(2x)=AC. We can easily find cos(x)=\sqrt{1-sin^2(x)). The answer is 253.

Let angle BCA be 3x, therefore, angle BAC is x.

From the sine law, we know that sin3x/504 = sinx/343. From the trigonometric identities, we know that sin3x = 3sinx - 4sin^3x.

Therefore (3sinx - 4sin^3x)/504 = sinx/343.

Expanding this we get sinx(sin^2x - 525/1372) = 0 Here, we have two cases, sinx=0 or sin^2x - 525/1372=0.

We know that sinx=0 is impossible because angle x would be 90 and the sum of the interior angles of the triangles will be larger than 180.

If sin^2x - 525/1372=0 then sinx = sqrt (525/1372) and x = arcsin (sqrt (525/1372).

Let angle ABC=y

To use the cosine law, we need to know angle y, and knowing that the sum of the angles in a triangle sums up to 180, we get the equation,

So,y=180-4(arcsin (sqrt (525/1372)))

Therefore, from the cosine law, we know that AC^2 = 504^2 + 343^2 - 2(504)(343)cosy

After subbing in y, we will get AC^2 = 64009

Therefore AC = 253

Suppose, ∠BAC = x then ∠BCA = 3x and since the sum of all angles of the triangle is 180 degress, therefore ∠ABC = (180-4x).

Now, applying the sine rule for a traingle:

sin x / 343 = sin 3x / 504 = sin (180-4x) / (length of AC) Solve the first two part of the equation to get the value of x and then solve the third part of the equation with any one of the other parts to determine the length of AC.

Calling angle BCA x, Using the law of sines and the expansions for sin 3x and sin 4x, which is equivalent to sin CBA, AC can be found with some computation.

Let $BC=a, CA=b, AB=c$ . By using sine rule, we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Since $C=3A$ , we have $\sin C=3\sin A-4\sin^3A$ . Hence, $504\sin A = 343 (3\sin A-4\sin^3A) \Leftrightarrow \sin A=\frac{5\sqrt{3}}{14}$ (since $0<\sin A<1$ ). Then $\sin B=\sin(180^\circ-4A)=\sin4A=4\sin A\cos A\cos 2A = \frac{1265\sqrt{3}}{4802}$ . Therefore, $b = \frac{343}{\frac{5\sqrt{3}}{14}} \cdot \frac{1265\sqrt3}{4802}=253$ .

Let $\angle BAC = \theta$ , so we have $\angle ACB = 3\theta$ . Place $D$ on $AB$ such that $\angle DCA = \angle DAC = \theta$ . Hence $DAC$ is isosceles and we have $\angle DCB = 3\theta - \theta = 2\theta$ . Also $\angle BDC = \angle DAC + \angle DCA = 2 \angle DAC = 2\theta$ . Hence $BDC$ is also isosceles.

Thus $DB=BC=343$ and $DC=DA=504-343=161$ . Applying the cosine rule to triangle $BDC$ , we have $\cos \angle ABC = \frac {343^2+343^2-161^2}{2 \times 343 \times 343}$ . Applying the cosine rule to triangle $ABC$ , we have $\cos \angle ABC = \frac {504^2 + 343^2 - AC^2}{2 \times 504 \times 343}$ . Equating these expressions, we solve for $AC$ (and use the fact that $504=343+161$ ) to obtain

$\begin{aligned} AC^2 & = (343+161)^2+343^2 +(161^2-2\times 343^2) \left(\frac {343+161}{343}\right) \\ & = 2\times 343^2 + 2 \times 343 \times 161 + 161^2 + 161^2 + \frac {161^3}{343} - 2 \times 343^2 - 2 \times 343 \times 161 \\ & = 2\times 161^2 + \frac {161^3} {343} \\ & = \frac {161^2 ( 2 \times 343 + 161) } {7^3} \\ & = 23^2 (121) \\ &= 253^2 \end{aligned}$

Therefore $AC = 253$ .

by sine law we get (3sinx-4sin^3x)/504=sinx/343 solving x from there then 180-4x .next we take cos of that value 2 343 504 and subtract it from 504 504+343*343. we take the squareroot of that no. and it is the value of AC.