Find the length X | Math Olympiad | A Very Nice Geometry Problem

Find the length X | Math Olympiad | A Very Nice Geometry Problem
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[@WhiteGandalfs]

That was a complicated road to Rome in the video! :D
I've gotten used to use mirror operations when two angles with 1:2 relation are given. Mirroring triangle ABD on AD to the other side of AC gives two nice rectangles which first give the division of AC into 4+6, then the length of AD as sqrt(45).

[@mariopopesco]

Angle ADB is @ and theta is noted with t. So angles ADC = 180 -@,
ACD = @ -t, ABD = 180 - (@+2t)
Law of sines in triangle ADC : sin t / 5 = sin (180- @) /10 -> sin @ = 2 sin t , so
cos @ = sqrt (1- 4 (sin t)^2)
Law of sines in triangle ABC : sine (3t) / (5+6) = sin (180 -(@+2t)) / 10
sin 3t = 3 sin t - 4 (sin t)^3
sin (180 - (@+2t)) = sin (@+2t) = sin @ cos 2t + sin 2t cos @ =
= 2 sin t (1 - (2 sin t)^2) + 2 sin t cos t sqrt ( 1 - 4 (sin t)^2) and cos t = sqrt (1- (sin t)^2) -->
[3 sin t - 4 (sin t)^3] /11 = [2 sin t (1 - 2 (sin t)^2) + 2 sin t * sqrt(1- (sin t)^2) * sqrt(1 -4(sin t)^2)] /10 --> divide by sin t and we have a equation with (sin t)^2; Let's say (sin t)^2 = y, we solve the quadratic equation and we have y = 1/5 and y = 35/32, but y = sin^2 so y (cos t)^2 = 4/5, and sin t = sqrt (1/5), cos t = sqrt (4/5)
Law of sines in triangle ADC : sin t /5 = sin (@- t) /x and
sin (@ - t) = sin @ cos t - sin t cos @ = 2 sin t cos t - sin t sqrt (1- 4 (sin t)^2) = 3/5
--> x = 3 sqrt5

[@prime423]

Much better explanation!!Kudos.

[@RAG981]

Lovely answer. Masterful.

[@juanalfaro7522]

There is a much faster way to solve the problem: Once AE is drawn, it bisects line BD into 2 equal parts, since triangles AEB and AED are congruent by ASA (BAE=DAE = theta, AE is common, and AED = AEB = 90). So BE = ED = 6/2 = 3. Once we find that AE/ED = AC/DC = 10/5 = 2, we have AE = 2*ED = 2*3=6. Then X^2 = AE^2 + ED^2 = 6^2 + 3^2 = 45 --> X = sqrt (45) = 3* sqrt (5)

[@skwest]

How do you know that:
∠AEB = ∠AED = 90°
?

[@maisonville7656]

@@skwest I was wondering the same. Not obvious to me that ABE and ADE are congruent. I did the whole calculation based on X^2=AB^2+AC^2-2*AB*AC*cos(theta), and taking cos(theta)=6/AB and find the same result. But again, I don't understand the basis for assuming ABE and ADE are congruent.

[@skwest]

@@maisonville7656
Yeah, I was hoping @juanalfaro7522 would clear up the mystery.

[@michaeldoerr5810]

Hey I was kind of wondering, does this technque apply always to triangles where the top angle is 2theta? And not only but does the x squared part qualify as a different construction?

[@bennyhsiao8435]

angleBAD=2
theta not theta

[@comdo777]

asnwer=5cm isit her

[@comdo777]

what whay 3/5 what asnwe=3/5

[@ALEducation-mz1jz]

1 min go for me...........😇