Germany Math Olympiad Problem | Best Math Olympiad Problem | Geometry

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Germany Math Olympiad Problem | Best Math Olympiad Problem | Geometry
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27 дек 2023

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Комментарии : 36

@timc5768 17 дней назад

A similar method might be: Let CF be perpendicular to AB, then (FB)/5 = 5/(2r), by similar triangles and radius = r. Then DC = 2r - 2(FB) = 2r - 25/r, so [(2r) - 25/r]/6 = (2r)/(6 + x) , sim triangles, [(DC)/(CE) = [(AB/(AC)], This gives: (2r^2 - 25)(x + 6) = 12r^2, (2r^2)x - 25x -150 = 0, or (4r^2)x = 300 + 50x , Eq. 1, and (2r)^2 = (6 + x)^2 + 5^2, Eq.2 , (pythag.),then by subst for 4r^2,from eq. 2 into eq. 1, we get x^3 + 12x^2 + 11x - 300 = 0

@math_qz_2 6 месяцев назад

Very instructive task

@LogicQuest 5 месяцев назад

I think you can reduce few steps here Y^2* (x+6)^2 = 900= 9.100= 3^2 * 10^2 [ other values does not satisfy due to lengths coming as not relevant] Y= 3 and X+6 =10 so X= 4 Cheers

@johnsavard7583 6 месяцев назад

Now this is a tough one. Since the arc CB is the same as the arc AD, we know the length of the line segment AD is also 5. If we call angle CAB theta, then angle DAB is also theta, while angle CAB is 90 degrees minus theta, so angle DAE is 2 theta minus 90 degrees. But where do I go from that? One possibility is: let's call the length of line segment DE by the name Y. Then we've got the equation 25 equals X^2 plus Y^2, and, in addition, Y^2 plus 36 is the square of the length of the line CD. Ah, now I see how an equation can be formed from which a solution might be found: since points C and D are both the same height above the diameter at the bottom, triangles ABC and CED are similar triangles! So 5 divided by 6 + X equals Y divided by 6, which can be substituted into the two equations involving squares for Y.

@zdrastvutye 6 месяцев назад

the first idea is to recognize the ensemble is symmetric, then i change the radius until the last equation fits: 10 l1=6:l2=5:nu=19:dim x(1,2),y(1,2):sw=(l1+l2)/1E2:r=l2/2+sw:goto 90 20 dis=4*r*r-l2^2:if dis0 then r1=r else r2=r 130 if abs(dg)>1E-10 then 120 140 if lx run using bbc basic sdl and hit ctrl tab to copy from the results window

@nantesloire 5 месяцев назад

Bei einem rechtseckigen Dreieck, grundsätzlich wenn der Hypertenuse 5 ist , dann sind die anderen Seiten nur 3 und 4 . Andere Möglichkeiten gibt es nicht.

@prime423 Месяц назад

We have seen this before. Extend DE to meet AB at F.We then have two parallel lines in a triangle. We can find DE in terms of C B and then use Pythagorean theorem. More time needs to be spent on analysis!!Anyone who can follow this knows about Pythagorean theorem and similiar triangles.

@dannyw66 Месяц назад

Nowhere in the initial problem does it state this is a semicircle so why assume it is? That could be erroneous and false

@zdrastvutye 2 месяца назад

the path to the result is: start lx with a small value and calculate the coordinate product ("skalarprodukt") because there is a 90degree angle in point E: 10 print "mathbooster-Germany Math Olympiad Problem":xa=0:ya=0 20 dim x(1,2),y(1,2):l1=5:l2=6:sw=l1/(l1+l2):lx=sw:lb=l1:n=l1^2+l2^2:goto 80 30 r=sqr((l2+lx)^2+l1^2)/2:la=l2+lx:lc=2*r:lh=(la^2-lb^2+lc^2)/2/lc 40 h=sqr(abs(la^2-lh^2)):xc=lh:yc=h:dxe=(xc-xa)*lx/(l2+lx):xe=xa+dxe:dye=(yc-ya)*lx/(l2+lx) 50 ye=ya+dye:yd=yc:xd=sqr(l1^2-yd^2):rem goto 58 60 dgu1=(xa-xe)*(xd-xe)/n:dgu2=(ya-ye)*(yd-ye)/n 70 dg=dgu1+dgu2:return 80 gosub 30 90 dg1=dg:lx1=lx:lx=lx+sw:lx2=lx:gosub 30:if dg1*dg>0 then 90 100 lx=(lx1+lx2)/2:gosub 30:if dg1*dg>0 then lx1=lx else lx2=lx 110 if abs(dg)>1E-10 then 100 120 print r,"%",lx :x(0,0)=0:y(0,0)=0:x(0,1)=2*r:y(0,1)=0:x(0,2)=xc:y(0,2)=yc 130 x(1,0)=0:y(1,0)=0:x(1,1)=xe:y(1,1)=ye:x(1,2)=xd:y(1,2)=yd 140 masx=1200/2/r:masy=900/r:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

@giuseppemalaguti435 6 месяцев назад

La relazione usata è arcsinx/5+arctg6/√(25-x^2)+arctg(x+6)/5=180,perché gli angoli opposti sono supplementari...applico tg e svolgendo i calcoli si arriva ad una quartica x^4+12x^3+11x^2-300x=0...che dà soluzione,oltre a x=0,x=4

@Emil_Avg 5 месяцев назад

Where do you find these problems?

@user-ws9zd8pn8r 4 месяца назад

I don't think the wording of this exercise is correct. I am launching a challenge, to draw the figure for X=4, it is not possible, try and tell me your opinions

@bizmyurt8582 5 месяцев назад

How can you know that d is mirrorpoint of c and that the big triangel is 90 degree, no symbol there?

@paranoidpolyglot 3 месяца назад

In big triangle ABC, angle C is an angle subtended in a semicircle, and there is a theorem of circles that states that any angle subtended in a semicircle is equal to 90°.

@zdrastvutye 2 месяца назад

wer hat denn da herumgepfuscht, dass man nicht als antwort schreiben kann z.b. (5/sqr(2))^2+(5/sqr(2))^2=25?

@bolder99 6 месяцев назад

Reason why Dc is parallel to AB needs to be clearly stated.

@tarunmnair 6 месяцев назад

because, AD=BC and ABCD is cyclic quadrilateral ( cyclic isoceles trapezium ).

@ludmilaivanova1603 6 месяцев назад

strictly saying, there should be mentioned that D and C are located at the same distance from the diameter therefore two lines are parallel.

@bolder99 6 месяцев назад

Cyclic isosceles quadrilateral covers the reason why DC is parallel to AB. By definition one pair of opposite sides has to be parallel and the other pair has to be equal.

@ludmilaivanova1603 6 месяцев назад

@@bolder99 how do we know that the quadrilateral is a trapezoid? how do we know it is cyclic?

@bolder99 6 месяцев назад

@@ludmilaivanova1603 That is what my initial comment was, the reason should be explained. The quadrilateral can be proved to be cyclic isosceles trapezoid.

@tanmayammont267 2 месяца назад

How to get 12 x

@ludmilaivanova1603 6 месяцев назад

is it not possible to solve the equation instead of guessing?

@jimlocke9320 6 месяцев назад

The third order equation y = x³ +12x² + 11x - 300 is continuous and will cross the x axis (y = 0) one time or three times. Use a half interval search to find one of the crossings. Find a value of x where y is negative and another where y is positive. A high enough negative value of x must produce a negative y, and a high enough positive value of x must produce a positive y. There must be 1 or 3 crossings in between. Add the 2 values of x, divide by 2 and compute y for this value of x. Take the 2 values of x where y is negative for one and positive for the other, add them, divide by 2 and compute y. Continue the process until y becomes very small (or zero). If x is very close to an integer and y is not 0, try the integer and see if you get an exact 0. In this equation, there is only one value of x where y is 0.

@ludmilaivanova1603 6 месяцев назад

@@jimlocke9320 Thank you. I have never solved equations this way. I was thinking of making this a product.I tried but did not get anywhere to solve it. This is my try: x(x^2+12x +6^2-6^2 +11)-300=0 x( (x+6)^2- 36+11) -300=0 x((x+6)^2 - 25) - 300=0 x(x+6-5)(x+6+5) =300 x(x+1)(x+11)=300 then just put 1,2,3,4 instead of x.

@tarunmnair 6 месяцев назад

yes, you can, but from rational root theorem, one/many of x = +/- 1,2,3,4,5,6,10,12,15,20,25,30,50,60,75,100,150,300; satisfies the equation. so by trying these values we can get atleast 1 value for this problem. This makes it easier to get the values of x. To solve for the cubic, x = {q + [q^2 + (r-p^2)^3]^1/2}^1/3 + {q - [q^2 + (r-p^2)^3]^1/2}^1/3 + p where p = -b/(3a), q = p^3 + (bc-3ad)/(6a^2), r = c/(3a) for a general form of ax3+bx2+cx+d=0; substitute the values and you get the same result.

@ludmilaivanova1603 6 месяцев назад

@@tarunmnair I saw the video with the solving a cubic equiation but I did not understand why there is a division by 3a

@nb9361 5 месяцев назад

If the hypotenuse of a right angled triangle is 5, this means it is a 3, 4, 5 triangle, therefore x=4.

@aavalos7760 4 месяца назад

Take a line segment of length 5 and rest one end anywhere on the x axis less than 5. If you knock that segment over, it will rotate and touch the y axis. We can slide it towards or away from the y axis, leading to many triangles satisfying a right angle triangle with hypotenuse 5.