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