Given that \(m \angle A= 60^\circ, BC=12\) units, \(\overline{BD} \perp \overline{AC}, \overline{CE} \perp \overline{AB}\) and \(m \angle DBC = 3m \angle ECB\), the length of segment \(EC\) can be expressed in the form \(a(\sqrt{b}+\sqrt{c})\) units where \(b\) and \(c\) have no perfect-square factors. What is the value of \(a+b+c?\)

VooFIX Nov 12, 2019

#1**+2 **

Call the center point X.

Quadrilateral AEXD has a total angle sum of 360 degrees.

Since XEA and XDA is 90 degrees, and angle A is 60 degrees.

This means that angle EXD is 90 + 90 + 60 + x = 360

240 + x = 360

x = 120

So angle EXD is 120 degrees.

Based on supplements, EXB is 60 degrees.

Triangles EXB and DXC are 30 60 90 triangles.

Can you find angles XBC and XCB?

Hint: you might need two variables and make a system of equations

CalculatorUser Nov 12, 2019

#3**+1 **

I got to a point where I basically knew most of the line segments but I'm trying to upload an image but it doesn't let me.

VooFIX
Nov 12, 2019

#4**+1 **

Since triangle DBC is a 45 - 45 - 90 triangle.

Can you solve for the length sof BD and DC?

CalculatorUser
Nov 12, 2019

#5**+1 **

Notice there are a lot of 90 degree triangles within the diagram,

I think you might have to create a system of equations that can solve for EC using pythagoream theorem knowing the length of DB and DC

CalculatorUser Nov 12, 2019

#6**+1 **

**I have finally managed to upload the photo! Here is the stuff that i have solved for.**

VooFIX Nov 12, 2019

#8**+1 **

DB and DC are congruent because of 45-45-90 so their length is \(6\sqrt2\). The image is still under moderation loading at 25% heres the link to Imgur if you don't feel like waiting for the moderation https://imgur.com/a/RLeBs91

VooFIX Nov 12, 2019

#9**+1 **

Find AD + DC,

We know the length of BD, so use 30 60 90 to find AD.

then use 30 60 90 triangle to solve for EC,

I think you got it!

CalculatorUser
Nov 12, 2019

#10**+1 **

AD should just be \(2\sqrt6\) so AD + DC is \(2\sqrt6+6\sqrt2\) but we don't have to do that step because AE is \(4\sqrt6\) since we know that BD is also \(6\sqrt2\) so EC is \(8\sqrt6\) ?

VooFIX Nov 12, 2019

#11**+1 **

I dont think AE is congruent to AD

Also, remember your answer is in the form \(a(\sqrt{b}+\sqrt{c})\)

CalculatorUser
Nov 12, 2019

#13**+1 **

Welp.

Do you have an AoPS class or something? Because if you submitted that answer then it is big OOOOFFF

I mean, at least you learned something!

CalculatorUser
Nov 12, 2019

#14**+1 **

Ye I have Aops, but its for alcumus and its apparently a level 24 geometry and I just decided to get stuck on this question. Oh ye, i didnt submit it yet so phew.

VooFIX
Nov 12, 2019

#15

#16**+2 **

Using the diagram by VooFIX and the Law of Sines we have that

12 / sin (90) = EC / sin (75)

So

12 sin (75) = EC

12 sin (45 + 30) = EC

12 ( sin 45 cos 30 + sin 30 cos 45) = EC

12 ( √2/2 * √3/2 + 1/2 * √2/2)

12 [ √6/4 + √2/4)

(12/4) ( √6 + √2) =

3 (√6 + √2)

So

a + b + c =

3 + 6 + 2 = 11

CPhill Nov 12, 2019

#17**+1 **

just a question, CPhill, was it possible to use 30 60 90 degree triangles to get the same answer?

CalculatorUser
Nov 12, 2019

#18**+2 **

I have the solution in terms of 30 60 90, but if I post it as a pic its gonna take forever for moderation so you want a link instead?

VooFIX
Nov 12, 2019

#19**+2 **

you can post the link.

the picture might work faster because CPhill might be here

CalculatorUser
Nov 12, 2019

#27**+2 **

Good spot, CU.....yeah....we can do this without Trig

Using VooFIX's diagram.......DC = 6√2 = √72

BD = DC because angles DBC and DCB are equal in right triangle BDC

And since PD is opposite a 30° angle in right triangle PDC....it has 1 / √3 the length of DC =

(1 / √3) * 6 √2 = 2√6 = √24

And

PC^2 = PD^2 + DC^2

PC^2 = 24 + 72

PC^2 = 96

PC = √96 = 4√6

BP = BD - BP = 6√2 - 2√6

And since EP is opposite a 30° angle in right triangle BPE....its length = (1/2) (BP) = 3√2 - √6

So

EC = PC + EP = 4√6 + ( 3√2 - √6) = 3√6 + 3√2 = 3 ( √6 + √2)

CPhill
Nov 12, 2019

#20

#22**+1 **

bruhhhhh lol at least my brain got its exercise in for today.

now time to do school homework

I have been postponing it because its way to easy and I feel like a peasant doing slave work when I do my homework.

CalculatorUser
Nov 12, 2019

#23**+1 **

XD So basically, you gain brain cells when you look at math forums and lose brain cells when you do homework.

VooFIX
Nov 12, 2019

#24**+2 **

I Gotta go too, I have all state auditions in 5 days and Jr. Honors Tom. so practicing while refreshing the page XD

VooFIX
Nov 12, 2019

#25**+2 **

yup, its a type of survival.

If I don't look at math forums, my IQ will constantly decrease until I am incapable of higher though.

This is to counteract the effects of the bad education system we have today in America.

Ok go practice practice oh life is hard

quote hamlet

to be or not to be

is life worth living?

is it worth the suffering?

we suffer because we are afraid of death.

CalculatorUser
Nov 12, 2019