Canada Math Olympiad Problem | A Very Nice Geometry Challenge
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At 5:17, Math Booster has constructed AE and found that it's length is the same as BE and CE. A perpendicular has been dropped from E to AB and the intersection designated F. Length EF has been designated x.
30 degrees
Using trigometry
In a 15, 75, 90 right triangle, the ratio of the sides are BC, sqrt 2 sqrt 2 (the hypotenuse), AB, sqrt 3 + 1 ( the longest base)
and AC, sqrt 3-1 (the shortest base)
Hence, let BC = 2n ( the hypotenuse),
Hence 2n = 2 sqrt 2
Hence AB in terms n = 2n/ 2 sqrt 2 * sqrt 3-1
= n/sqrt 2 * sqrt 3-1 divide numerator and denominator by 2
= sqrt 2 n /2 * sqrt 3-1 get rid of the radical from the bottom
=(sqrt 6- sqrt 2)/2 n
= 0.51763 n
AB = (sqrt 6 - sqrt 2/2)n or 0.51763 n
Since BC = 2n, then EC = 1 n (since it is one-half BC)
Draw a line from A to E to form a triangle ADE and ACE.
to find length A E, let's use the Law of Consine
no need to include the 'n'
AE = sqrt (EC^2 + AB^2 - 2 EC AB cos 75 degrees)
AC = sqrt (1^2 + 0.51763^2 - 2*1*0.51763. 0.2588190451
sqrt ( 1 + 2679408169 - 2*0.2588190451*v 0.51763) the 2 times cos 75 = 0.51763. that's why they cancel out
= sqrt (1 + 2679408169 - 2679408169)
= sqrt 1
= 1
Hence, triangle ACE is an isosceles triangle since AC = EC (recall EC= 1n , and now AC = 1n)
Since angle C = 75 degrees (ABC is a 15, 75, 90 triangle), angle A (in ABC or BAC) = 75 degrees.
Hence, triangle ADE is also isosceles 15 degrees ( 90 -75), 15 degrees, and 150 since AD = DE (given in video)
Let's find the length of AD using the Law of Sine
1/ sin 150 degree = AD/sin 15 degree
1* sin 15 degrees/ sin 150 degrees = AD
=0.51763
Hence AD = AC
And since triangle ACD is a right triangle ( see video)
then it is a 45, 45, 90-degree right triangle; therefore, angle ACD is 45 degrees; therefore, theta = 30 (75 - 45). Answer
When I found out that AD = AC, I was surprised as AC looks greater than AD 10:00
Constructing EF perpendicular AB is the 'Key' step.Exellent.
Fine. Greetings from Spain. I am fan of your Channel
Since ∆AEC is Isosceles
and ∠ACE =75°, ∠AEC=30°
Reflect ∆EDA about EA to form ∆ED'A. ∆ED'C is Congruent to ∆EDA by SAS. Hence ∠AD'C=60°, and since AD' =CD', ∆AD'C is Equilateral.
This implies AC = AD and therefore ∠ACD=45° and θ= 30°.
when you dropped a perpendicular on AB from the middle point of BC, you created a middle line of a triangle which is eqaual to a half of the base (AC). Sonce at the same time EF is equal to a half of DE it means AC=DE=DA.
The answer is 45 degrees. Golly I am just AMAZED how having to use the exterior angle theorem because the third angle is more than 90 degrees while the angles are the same leads up to similar triangles. Looks like I NEED to collate all of the similar triangle problems on your channel and test them out so that I am easily baffled.
I'm surprised you got the answer right! You're missing a lot of math, mate! You simply have to study a lot more than you think!
@@imetroangola4943 Hey. I am actually catching up on almost all of the math that I could be learning. This is just one of the channels that I have been looking up on. I have also been commenting and watching Math Booster videos as well as Dr. Yap Keem SIAH videos. I am planning on mixing and matching more and more math problems so that I can remain at peak elite mathematician mode. Or JEE intuition.
{15°B+15°C+90°A}= 120°BCA {30°D+30°E}=60°DE {120°BCA+60°DE}= 180°BCADE 60^60^60 2^30^2^30^2^30 1^5^6^1^5^6^1^5^6 1^3^2^1^3^2^1^3^2 1^1^1^1^3^2 3^2 (BCADE ➖ 3BCADE+2).
30degrees