Trigonometric Lengths

Visit Brilliant's blog on Cosine Rule or Wikipedia - Apollonius's Theorem for more details and proof about Apollonius's Theorem.

Solution of the Problem:

Consider triangle $DCB$ ,
$CE$ is a median.
Therefore, Applying Apollonius's Theorem on triangle $DCB$
$BC^{2} + CD^{2} = 2(DE^{2} + CE^{2})$
Similarly in triangle $ACE$ ,
$CD$ is a median.
Therefore, Applying Apollonius's Theorem on triangle $ACE$
$AC^{2} + CE^{2} = 2(DE^{2} + CD^{2})$
Adding the above two relations yields,
$BC^{2} + CD^{2} + AC^{2} + CE^{2} = 2CE^{2} + 4DE^{2} + 2CD^{2}$
But $AC^{2} + BC^{2} = AB^{2}$ (Applying Pythagorean Theorem in triangle $ABC$ ). Using this relation and simplifying the above relation by cancelling common terms both the sides we get
$AB^{2} = CE^{2} + CD^{2} + 4DE^{2}$
Now as it is given that
$CD = 5\cos\theta$ and $CE = 5\sin\theta$ Our equation simplifies to
$AB^{2} = 25\sin^{2}\theta + 25\cos^{2}\theta + 4DE^{2}$
And as $\sin^{2}\theta + \cos^{2}\theta = 1$
$AB^{2} = 25 + 4DE^{2}$
But we know $DE = \frac{AB}{3}$ Therefore
$AB^{2} = 25 + \frac{4AB^{2}}{9}$
Hence
$(1 - \frac{4}{9})AB^{2} = 25$
And this gives $AB^{2} = 45$

First, we can draw 4 line segments from points $D$ and $E$ to side $CB$ and $AC$ such that these 4 segments are perpendicular to sides $CB$ and $AC$ . Since points $D$ and $E$ are evenly spaced on segment $AB$ , we know that the perpendicular line segments drawn in must divide $CB$ and $AC$ each into 3 segments of equal length.

With these new segments, we have new right triangles. We now know that $(\frac {2}{3}CB)^2 + (\frac {1}{3}AC)^2 = (5\sin \theta)^2$ and $(\frac {2}{3}AC)^2 + (\frac {1}{3}CB)^2 = (5\cos \theta) ^2$ .

If we combine these two equations by adding them together, we get $(\frac {2}{3}CB)^2 + (\frac {1}{3}AC)^2 + (\frac {2}{3}AC)^2 + (\frac {1}{3}CB)^2 = (5\sin \theta)^2 + (5\cos \theta)^2$ which can be simplified to $\frac {5}{9}AC^2 + \frac {5}{9}CB^2 = 25\cos \theta^2 + 25\sin \theta^2$ .

We can then factor out the 25 from the right side of the equation. We know our trig identity which states that $\sin \theta^2 + \cos \theta^2 = 1$ , so we can substitute it with one, leaving us with $\frac {5}{9}AC^2 + \frac {5}{9}CB^2 = 25$ . After further simplification, we are left with the equation $AC^2$ + $BC^2$ = 45 .

In this equation, we are asked to find $AB^2$ . Since it is a right triangle, we know that $AC^2$ + $BC^2$ = $AB^2$ .

Therefore $AB^2 = 45$ .

Drop the perpendiculars from D and E to BC, called F and G respectively.

Let $\alpha = \angle DCF.$ Let $\beta = \angle CEG.$

Then $DF = \frac{1}{3}AC = \sin\alpha5\cos\theta$ $CG = \frac{1}{3}BC = \sin\beta5\sin\theta$

$\frac{1}{9}AB^2 = 25(\sin^2\alpha\cos^2\theta + \sin^2\beta\sin^2\theta)$

$CF = \frac{2}{3}BC = \cos\alpha5\cos\theta$ $GE = \frac{2}{3}AC = \cos\beta5\sin\theta$

$\frac{4}{9}AB^2 = 25(\cos^2\alpha\cos^2\theta+\cos^2\beta\sin^2\theta$

$\frac{5}{9}AB^2 = 25(\sin^2\alpha\cos^2\theta + \sin^2\beta\sin^2\theta + \cos^2\alpha\cos^2\theta + \cos^2\beta\sin^2\theta)$

$= 25(\cos^2\theta + \sin^2\theta) = 25$ $25 \times \frac{9}{5} = 45$

Let $∠CAB=\alpha$ and $∠ABC=90-\alpha$ we have that $CD^2= AC^2 + AD^2 -2AD \cdot AC \cos\alpha$ and $CE^2=EB^2+CB^2-2EB\cdot CB \cos(90-\alpha)$ due to cosine rule. So we have $CD^2= AC^2 + DE^2 -2DE \cdot AC \cos\alpha$ and $CE^2=DE^2+CB^2-2DE\cdot CB \cos(90-\alpha)$ since $AD=DE=EB$ . We obtain $CD^2+CE^2=AC^2+CB^2 +2DE^2 \\ -2DE(AC \cos\alpha+ CB \cos(90-\alpha))$

by summing the equations above. Now we have that $CD^2+CE^2=25((\cos\theta)^2+(\sin\theta)^2)=25$ ( since $(\cos\theta)^2+(\sin\theta)^2=1$ ). $AC^2+CB^2=AB^2=9DE^2$ (Pythagorean theorem) and $AC \cos\alpha+CB\cos(90-\alpha)=AB=3DE$ . So we have $25=9DE^2+2DE^2-2DE\cdot(3DE)$ and so $25= 5DE^2$ by substituting in the above equation. So $DE^2=5$ and $AB^2=9DE^2=45$ .

In triangle AEC , CD is the median to AE. apply apollonius equation to it. similarly in triangle BCD, CE is the median to BD. apply apollonius equation to it. let AE=DE=BE=x and CD^2 +CE^2=25 and AC^2 +BC^2=9x^2. solve for x from the two equations.

We are told $AD = DE = BE$ , therefore D and E must lie between A and B. For my explanation I will assume they lie in the order $ADEB$ . If D and E are swapped, the results are the same. We can therefore deduce $AE = 2AD$ and $AB = 3AD$ .

We are told $CD = 5cos\theta$ and $CE = 5 sin\theta$ so $CD^2+CE^2 = 25$ by the Pythagorean Identity

As $ACB$ is a right triangle, we know $AC = ABcosA = 3ADcosA$

By the cosine rule:

$CD^2 = AC^2 + AD^2 - 2AC\cdot AD cosA$ $CE^2 = AC^2 + AE^2 - 2AC\cdot AE cosA$

therefore $CE^2 = AC^2 + 4AD^2 - 4AC \cdot AD cosA$

By adding the first and third equations we get $25 = 2AC^2 + 5AD^2 - 6AC \cdot AD cosA$

so

$25 = 2(3ADcosA)^2 + 5AD^2 - 6(3ADcosA)ADcosA$

$25 = 18AD^2 cos^2 A + 5AD^2 - 18AD^2 cos^2 A$

$25 = 5 AD^2$

$AD = \sqrt(5)$

$AB = 3 \sqrt(5)$

$AB^2 = 45$

Assume a coordinate system with the origin in the point $C$ and with points $A = (b,0)$ and $B = (0,a)$ for some arbitrary $a$ and $b$ .

The length of the hypotenuse $AB$ is then equal to $\sqrt{a^2 + b^2}$ .

Since $BE = ED = DA = \frac{1}{3} AC$ it is not hard to deduce the coordinates of points $D$ and $E$ . Think of a rectangle with corners in $A$ , $B$ and $C$ and divide it with a 3 by 3 grid.

Precisely, $D = (\frac{2}{3}b,\frac{1}{3}a)$ and $E = (\frac{1}{3}b,\frac{2}{3}a)$ .

We can then calculate $CD$ and $CE$ easily, since $C$ is the origin.

$CD = \sqrt{\frac{4}{9}b^2 + \frac{1}{9}a^2} , CE = \sqrt{\frac{1}{9}b^2 + \frac{4}{9}a^2}$

We also have $CD = 5 \sin\theta , CE = 5 \cos\theta$ .

After we have eliminated $CD$ and $CE$ , squared both equations and added them together we have:

$\frac{5}{9}b^2 + \frac{5}{9}a^2 = 25(\sin^2 \theta + \cos^2 \theta)$

Continuing on, we obtain:

$\frac{5}{9}b^2 + \frac{5}{9}a^2 = 25$

$\frac{1}{9}b^2 + \frac{1}{9}a^2 = 5$

$b^2 + a^2 = 45$

Notice that $AB^2$ is actually equal to $b^2 + a^2 = 45$ , and so the solution is 45.

By Stewart's Theorem applied to $\triangle ACE$ , and by the facts $AE = 2AD$ and $AB^{2} = 9AD^{2} = AC^{2} + BC^{2}$

$AD(25sin^{2}\theta + AC^{2}) = 2AD(25cos^{2}\theta + AD^{2})$ $\Rightarrow 25sin^{2}\theta + AC^{2} = 50cos^{2}\theta + 2AD^{2}$

Similarly, applying Stewart's Theorem to $\triangle CBD$ , $25cos^{2}\theta + BC^{2} = 50sin^{2}\theta + 2AD^{2}$ .

Adding the two equations, rearranging, and using $sin^{2}\theta + cos^{2}\theta = 1, AB^{2} = 9AD^{2} = AC^{2} + BC^{2}$ ,

$5AD^{2} = 25 \Rightarrow AD^{2} = 5 \Rightarrow 9AD^{2} = 45 \Rightarrow AB^{2} = 45$ .

Denote length of DE = x = AD = BE.

Apply Stewart's Theorem to triangle ACE and triangle BCD.

We have two equations:

xCE^2 + xAC^2 = 2x^3 + 2xCD^2 xCD^2 + xBC^2 = 2x^3 + 2xCE^2

Add the two equations. Also note that CD^2 + CE^2 = 25 and AC^2 + BC^2 = AB^2. Hence we get

25x + xAB^2 = 4x^3 + 50x AB^2 = 4x^2 + 25

AB^2 = (AD + DE + EB)^2 = (3x)^2 = 9x^2

9x^2 = 4x^2 + 25, therefore x^2 = 5. AB^2 = 9x^2 = 45.

By property of all right triangles, we will have (5\cos \theta)^2 + (5\sin\ theta)^2 = (5/9)(AB)^2 25\cos^2 \theta + 25\sin^2 \theta = (5/9)(AB)^2 25(\sin^2 \theta + \cos^2 \theta) = (5/9)(AB)^2 25(9/5) = (AB)^2 45 = (AB)^2

Applying Stewart's Theorem, we get AC^2(BD) +BC^2(AD) = AB((BD)(AD)+CD^2) and AC^2(BE) +BC^2(AE) = AB((BE)(AE)+CE^2) since AD=DE=BE, we add both equations and express everything in terms of AD, we get AC^2(3AD)+ BC^2(3AD)=3AD(2AD^2+25(cos^theta +sin^2 theta)) using AC^2 + BC^2 = AB^2 and cos^theta +sin^2 theta = 1, after factoring 3AD we get AB^2 = 4AD^2+25, using the fact that AB = 3AD and solving the equation, we get AB^2 = 45.

Applying Apollonius’ theorem ( Worked Example 3 ) to triangles $ACE$ and $DCB$ , we have

$AC^2 + 25 \sin ^2 \theta = 2 \left(25 \cos^2 \theta + \frac {1}{9} AB^2\right),$ $BC^2 + 25 \cos^2 \theta = 2 \left( 25 \sin^2 \theta + \frac {1}{9} AB^2 \right) .$

Summing the 2 equations, we obtain $AC^2 +BC^2+ 25 = 50 + \frac {4}{9} AB^2$ . By the Pythagorean theorem, we have $AC^2 + BC^2= AB^2$ , hence $AB^2 = 25 \times \frac {9}{5} = 45$ .

Draw straight lines through E and D parallel to BC, and through E and D parallel to AC, And let the distance form D to AC is x, and form E to BC is y. So we have AB^2 = 9 AD^2 = 9(x^2 +y^2). We have also CD^2 + CE^2 = 25, (x^2 +4 y^2) + (4 x^2 +y^2) = 25, or (x^2 +y^2)=5 . So AB^2 = 45

In $\triangle ACD$ ,

$cosA = \frac{AC^2 + AD^2 - CD^2}{2*AC*AD} = \frac{AC}{AB}$

$\frac{AC^2 + DE^2 - 25cos^2\theta}{2*AC*DE} = \frac{AC}{3DE}$

$\boxed{AC^2 + DE^2 - 25cos^2\theta = \frac{2}{3} AC^2}$

In $\triangle BCE$ ,

$cosB = \frac{BC^2 + BE^2 - CE^2}{2*BC*BE} = \frac{BC}{AB}$

$\frac{BC^2 + DE^2 - 25sin^2\theta}{2*BC*DE} = \frac{BC}{3DE}$

$\boxed{BC^2 + DE^2 - 25sin^2\theta = \frac{2}{3} BC^2}$

Adding the 2 equations,

$AB^2 + 2DE^2 - 25(1) = \frac{2}{3} AB^2$

$AB^2 + 2(\frac{AB^2}{9}) - \frac{2}{3} AB^2 = 25$

$\boxed{AB^2 = 45}$

in∆ cdb e is mid point of bd so ce is perpendicular to bd & CD=cb=5cos(theta). cos(cbd)=5cos(theta)\3be at∆ ABC. cos(cbd)=be\cos(theta),,cos^2(theta)\be^2=3\25. be=5cos(theta)\√3.at∆cbe cos(cbe)=5cos(theta)\√3÷5cos(theta) so. cos(cbe)=1\√3. So tan(cbe)=√2=5sin(theta)(5cos(theta)\√3). so tan(theta)=√2\√3. cos(theta)=√3\√5. So be=(5\√3)×(√3\√5)=√5. AB=3be=3√5 so AB^2=9×5=45########

We first may assume the angle is 45 degrees (since this is AIME), implying that CD=CE=5rt2/2. Now since CD=CE, it is implied that ABC is isosceles. Now we say AD=DE=EB=x. Therefore BC=3x/rt2. Apply law of cosines on CEB, we have (5rt2)^2=(3x/rt2)^2+x^2-2•(3x/rt2)•xcos45. Solving, we find x^2=5. We are looking for 9x^2, so our answer is 45.