Distances in a Right Triangle

Applying the Pythagorean theorem on triangle $ABC$ , we have $AC = \sqrt{AB^2 + BC^2} = \sqrt{(30\sqrt{3})^2 + (30)^2} = 60$ . Thus $CBA$ is a $30^\circ - 60^\circ - 90^\circ$ triangle as $CB: BA: AC = 1: \sqrt{3}: 2$ . Therefore $\angle ACB = 60^\circ$ and $\angle BAC = 30^\circ$ . This implies that $\angle EBC = 30^\circ$ and so triangle $BEC$ is a $30^\circ - 60^\circ - 90^\circ$ triangle. Hence $EC = 15$ and $BE = 15\sqrt{3}$ . Thus it follows that $AE = AC - EC = 60 - 15 = 45$ .

From $D$ drop a perpendicular to $AC$ at $G$ . So, we have that triangle $DGC$ is a $30^\circ - 60^\circ - 90^\circ$ triangle, thus $DG = \frac{15\sqrt{3}}{2}$ and $GC = \frac{15}{2}$ . Hence $AG = AC - GC = 60 - \frac{15}{2} = \frac{105}{2}$ . We also have that triangles $AEF$ and $AGD$ are similar by angle-angle-angle. Thus $\frac{EF}{AE} = \frac{DG}{AG} \Rightarrow EF = \frac{AE\cdot DG}{AG} = \frac{45 \cdot \frac{15\sqrt{3}}{2}}{\frac{105}{2}} = \frac{45\sqrt{3}}{7}$ . Hence $a + b + c = 45 + 3 + 7 = 55$ .

Key Ideas :

(1). Form equation of lines AD and BF and then find their point of intersection E.

(2). Using distance formula find EF.

Calculations:

To form the equation of straight line AD using two point formula- Take BC along x axis and B as origin, find points A and D. It will come out to be A (0, 18 $\sqrt{3}$ ) and D (9,0).

To form the equation of straight line BF - If you know how to find the foot of a perpendicular from a point on a straight line find point F and hence equation BF

Otherwise, use the the slope point formula to derive the equation of BF. Note that to find the slope of line BF you can use the fact that BF is perpendicular to AC.

After finding the equations of both the lines find their point of intersection E.

Finally, using the distance formula find distance between the point E and F and hence you have EF which is required in the question.

We find $BE$ first using area of the triangle. $\frac{1}{2} \times BC \times AB = \frac{1}{2} \times AC \times BE$

Using $AC = \sqrt{AB^2 + BC^2}$ , we get $BE = 15 \sqrt{3}$

Now, $\tan(x) = \frac{AB}{BD} = 2 \sqrt{3}$ . So, $\sin(x) = \frac{2 \sqrt{3}}{ \sqrt{13}}$ and $\cos(x) = \frac{1}{ \sqrt{13}}$

Using sine rule in $\Delta BFD$ , we get $\frac{\sin(150^o -x)}{15} = \frac{\sin(x)}{BF}$ which upon expanding and solving gives us $BF = \frac{60 \sqrt{3}}{7}$

Finally, $EF = BE - BF = \frac{45 \sqrt{3}}{7}$

And, of course, $45+7+3=55$

AC=√[(30√3)²+30²]=60

From similarity, AE=45 and EC=15

Also in triangle ABD, AD=√[(30√3)²+15²]=√2925=15√13

Applying cosine rule in traingle ADC (taking angle DAC=∅)

cos∅= (AD²+AC²-CD²)/(2xADxAC)

cos∅= (2925+3600-225)/(2x15√13x60)=7/2√13

Which gives sin∅=√3/2√13 and tan∅=√3/7

In triangle AEF, right angled at E,

EF=AE tan∅=45√3/7

Hence a+b+c=45+3+7=55

Relevant wiki: Menelaus' Theorem

In $\Delta ABC$ ,

By Pythagorus theorum, we have :

$AC=\sqrt{(30\sqrt3)^2+30^2}=60$

In $\Delta ABC\text{ and }\Delta BEC$ we observe:

$\angle B=\angle E=90^{\circ}$

$\angle C=\angle C$

Therefore $\Delta ABC\sim\Delta BEC$

$\Rightarrow EC=\frac{30}{2}\text{ and }BE=15\sqrt3$ so $AE=60-15=45$

Now, In $\Delta BEC,\text{ } D,F,A$ are externally colinear. So by Menelaus' Theorem ,

$\frac{BD}{DC}\times\frac{CA}{AE}\times\frac{EF}{FB}=1$

$\frac{15}{15}\times\frac{60}{45}\times\frac{EF}{FB}=1$

$\frac{EF}{FB}=\frac{3}{4}$

and we also have $EB=EF+FB=15\sqrt3$

$\Rightarrow EF=45\frac{\sqrt3}{7}$

So the answer is $\boxed{55}.$

Simple calculations yield $AE = 27, BE = 9\sqrt{3}$ , and CE = 9. Thus AE/CE = 3, and BD/DC = 1. We assign a mass of 1 to A, so B and C get masses of 3, E has a mass of 4, and D has a mass of 6. Thus EF/FB = 3/4, so EF/EB = 3/7. Thus $EF = 3/7 \times 9 \sqrt{3} = 27\sqrt{3}/7$ , so the answer is 37.

[Note: There wasn't a need to calculate the mass at D.]

First, we use the trig operator $\tan$ and $\cos$ respectively over $\angle BAE$ to find that $\angle BAE = 30 ^ \circ$ and $AE=45$ .

Now we let $\angle EAF = \theta$ which implies that $\angle BAD = 30 ^ \circ - \theta$ . Therefore, $\tan \ (30 ^ \circ - \theta) = \frac {BD}{AD} = \frac {\sqrt{3}}{6}$ .

Using the angle subtraction formula for tangent ,we solve that, $\tan \theta = \frac {\sqrt{3}}{7}$ . Since $\bigtriangleup AEF$ is also right angled we conclude that, $EF = AE \times \tan \theta = \frac {45\sqrt{3}}{7}$ and hance the ans is $45 + 3 + 7 = \boxed{55}$