Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods

Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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[@SuperPassek]

Interesting solutions. Another solution: Let H be a point on BC s.t. DH ⊥ BC. Then CDH = 75 and DH = (1/2)AB. As tan 75 = 2 + √3, HC = (2 + √3) DH. From CE = 2 DH, we get HE = √3 DH, which means EDH = 60. So EDC = 15

[@giuseppemalaguti435]

ctgθ=1/2(sin15)^2-ctg15=2+√3...θ=15

[@jimlocke9320]

As in the video, let length AB = CE = b and AD = CD = a, therefore CA = 2a. Construct a line segment AF such that F is on BC and

[@ducduypham7264]

From B draw a line perpendicular to AC that intersect AC at H. Because ABC is right triangle with hypotenuse AC and angle BAC=75° so AC=4BH. Additionally AC=2DC as D is midpoint of AC, then DC=2BH. Construct point B' so that H is midpoint of BB'. Triangle ABB' congruent to triangle ECD (side-angle-side) as AB=EC, angle ABB'=angle ECD=15° (because both angle ABB' and angle DCE are complementary angles of angle HBC), BB'=DC=2BH. Therefore theta=angle EDC=angle AB'B. We can easily prove that triangle ABB' is isosceles triangle with base BB' so angle AB'B=angle ABB'=15°. In conclusion theta=15°

[@Irtsak]

Let BD the median in triangle ABC and EP⊥AC (construction) Let AD=DC=x and AB=EC=y In orthogonal triangle ABC, BD is median => BD=AD=DC =x => triangle ABD is isosceles => ∠BAD= ∠ABD =75°. So ∠ADB=30° When the angle of a right triangle is equal to 30°, remember that the length of opposite side is always equal to half of the length of the hypotenuse. => *AE=x/2* (1) Orthogonal triangles ABE=EPC (cause AB=EC=y and ∠ABE= ∠PEC=75°) So PC=AE => PC=x/2 cause (1) Although DP=DC-PC=x-x/2=x/2 => DP=x/2 .

[@Ml-sy6uo]

We can have AM so that M is on BC and

[@user-yz8ex1rr1i]

Достраиваем до квадрата со стороной равной ВС. А потом внутри квадрата строим равносторонний треугольник со стороной равной стороне квадрата с вершиной на точке D и дальше решается очен просто.

[@alexbayan8302]

Draw a line AF so that the angle BAF is 60 degree. If AB is a, AF is 2a and FC is 2a. Then FE=CE. Since AD=CD too, triangles AFC and DFC are similar. So angle CDE is the same as FAC. Since the triangle FAC is isosc, angles CDE = FAC = FCA = 15 degree.

[@alexbayan8302]

I wrote a lot to explain the reasoning process but an experienced solver will see the relation instantly. Just draw AF and the rest is seen instantly.

[@michaeldoerr5810]

I thought that the first method kind showed how trig identifies can be used in relation to the cotangent function. Also for the second method, if there was a congruence postulate for the two pairs of congruent triangles, it would be SAS. I could be wrong and I shall make this practice for geometry.

[@sinsn2147]

Let P on BE so that PAB=60 deg Thus, AP=2AB Also, PAC=PCA=15 deg, so AP=PC=2AB=CE+EP=AB+EP => EP=AB CD/CE=CA/CP, finding that ΔCED is similar to ΔCPA EDC=PAC= 15 deg

[@arulbiswas1260]

using cos(a-b) formula would have been much easier.

[@devondevon4366]

15 degrees

[@comdo777]

asnwer=15 isit

[@professorrogeriocesar]

Difícil.