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

Из точки D,описываем окружность, из точки С, на окружности засекаем точку. отстоящую от С на расстоянии АВ=СЕ, и соединяем эту точку К, с точкой D, Получим треугольник CKD, который равен треугольнику АВD, DE продлеваем до пересечения с СК в точке М, и видим, что DM является биссектрисой, медианой и высотой треугольника CKD, и угол тета, равен половине угла АВD, Который равен углу СDK, и равен 15 градусов.

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.

∠ECD =15°. Let's take EC=1. Then AD=BD=CD=0.5csc15°. In ΔCDE, we know that EC/sinθ = CD/sin(θ+15°), i.e. 1/sinθ = 0.5csc15°/sin(θ+15°) = 2cos15°/sin(θ+15°). Here I change 0.5 to sin30°, and sin30°csc15°=2cos15°. Then we get sin(θ+15°)/sinθ=2cos15°. Because sin(θ+15°)=sinθcos15°+cosθsin15°, sin(θ+15°)/sinθ=cos15°+cotθsin15° = 2cos15°. Finally, we get cotθ = cos15°/sin15° = cot15°, i.e. θ = 15°.

Just draw a line from A to F s.t. BAF is 60 deg. It is simple to see that FDC is a right angle triangle in a circle with diameter FC and center at E. That means ED=EC and angle theta = angle C = 15 deg.

ctgθ=1/2(sin15)^2-ctg15=2+√3...θ=15

to find the angles, draw a line in D that is parallel to line BC: 10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu0:gcol8:@zoom%=@zoom%*1.4 20 print "mathbooster-russian math olympiad-a very nice geometry problem" 30 dim x(1,2),y(1,2):la=1:w1=75: sw=.1:wth=sw:goto 60 40 w2=90-w1:w3=wth+w2:w4=180-w3:lbu=la*sin(rad(wth))/sin(rad(w4)) 50 l2=2*la*cos(rad(w1)):dg=l2/lbu:dg=dg-1:return 60 gosub 40 70 dg1=dg:wth1=wth:wth=wth+sw:gosub 40:wth2=wth:if dg1*dg>0 then 70 80 wth=(wth1+wth2)/2:gosub 40:if dg1*dg>0 then wth1=wth else wth2=wth 90 if abs(dg)>1E-10 then 80 100 print "der gesuchte winkel="; wth;"°" 110 x(0,0)=0:y(0,0)=0:x(0,1)=2*la*sin(rad(w1)):y(0,1)=0:x(0,2)=0:y(0,2)=l2 120 x(1,0)=x(0,1)-l2:y(1,0)=0:x(1,1)=x(0,1):y(1,1)=0:x(1,2)=x(0,1)-la*cos(rad(w2)) 130 y(1,2)=la*sin(rad(w2)):masx=1200/x(0,1):masy=850/l2 140 if masxathbooster-russian math olympiad-a very nice geometry problem der gesuchte winkel=15° run in bbc basic sdl and hit ctrl tab to copy from the results window

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

We can have AM so that M is on BC and

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 .

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°

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.

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.

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

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

15 degrees

Difícil.

asnwer=15 isit