Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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- เผยแพร่เมื่อ 16 พ.ค. 2024
- Russian Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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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
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.
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.
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
ctgθ=1/2(sin15)^2-ctg15=2+√3...θ=15
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°
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 .
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.
We can have AM so that M is on BC and
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.
using cos(a-b) formula would have been much easier.
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
15 degrees
Difícil.
asnwer=15 isit