Germany Math Olympiad Problem | A Very Nice Geometry Challenge

Germany Math Olympiad Problem | A Very Nice Geometry Challenge

MY OTHER CHANNELS
••••••••••••••••••••••••••••••••

--------------------------------------------------------------------------------

Become a member of this channel to get access of ULTIMATE MATH COURSE
(New video will be uploaded every week)

Join the channel to become a member

[WahranRai]

Another approach would be to use the Cartesian equation of the tangent through C to the circle and express that the distance from the center O of the circle to the tangent is equal to the radius.
With axis X = AB and axis Y = D, center O = (1, 1), C = (4, 4)
Equation of the tangent passing thru C : ax + by + c = 0 with y-4 = m(x-4) ---> mx-y-4m+4 = 0 --->
a = m, b=1, c = -4m+4
Distance of O to tangent : abs(ax_O + by_O +c) / sqrt(a^2 + b^2) ....

[michaellawson]

The problem should of stated from the beginning that the diagonal of the square passes through the center of the circle. That fact allows the rest of the solution to flow. Putting a ruler up to the square leads proves that the center does not line up with the diagonal. Frustrating to say the least.

[musluktandokulenkarabatak]

Another approach is to use angles. When you draw to perpendiculars from center to AB and CE, you get a deltoid. The arc inside the deltoid equals to angle CEB, and since the angle of deltoid's which is at the center of the circle sees that arc, it equals CEB. When you draw a line from center to E, you get an angle bisector so new angles are CEB/2. Then you can use tangent half angle formula in order to calculate x.

[abdmoh]

Thank you so much for this great exlanation.

[ActualDumBatcha]

this was on my year 8 maths test, except we were finding the hypotenuse, not the area. i remember no one knowing how to do this as the last question, and trying to do this for the next few nights. super traumatic.

[zdrastvutye]

for every tangent there is a thales theorem:
10 dim x(3), y(3):l1=4:r=1:sw=.1:xm=r:ym=r:xs=xm:rem den thalessatz anwenden
20 50
30 disy=rt^2-(xs-xmt)^2:if disy<0 then return
40
50 gosub 30:if disy<0 then else 70
60 xs=xs+sw:goto 50
70 dg1=dg:xs1=xs:xs=xs+sw:if xs>10*l1 then stop
80 xs2=xs:gosub 30:if dg1*dg>0 then 70
90 xs=(xs1+xs2)/2:gosub 30:if dg1*dg>0 then xs1=xs else xs2=xs
100 if abs(dg)>1E-10 then 90
110 print xs, "%", ys
120 xe:ages=(l1-xe)*l1/2:print "die flaeche=";ages
130
140 mass=8E2/l1:goto 160
150 xbu=x*mass:ybu=y*mass:return
160 xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0
170 x=x(ia):y=y(ia):gosub 150:xbn=xbu:ybn=ybu:goto 190
180 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
190 gosub 180:next a:x=xm:y=ym:gcol5:gosub 150:circle xbu, ybu, r*mass
200 x=xe:y=0:gosub 150:xbn=xbu:ybn=ybu
210 gcol4:gosub 180
1.85385094% 0.479482396
1.56155281
die flaeche=4.87689437
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[alexandertokarev]

EO - биссектриса угла AEC,
поэтому CE/AE = CO/AO = 3.
Пусть z = BE<4, по теореме Пифагора для EBC:
CE^2 = z^2+BC^2 = z^2+16 = (3•AE)^2 = 9•(4-z)^2.
Откуда олучается квадратное уравнение с неизвестной z:
z^2-9•z+16=0.
Корни 9/2+√17/2 и 9/2-√17/2.
Только второй корень <4, значит
[EBC] = z•4/2 = 9-√17.

[quigonkenny]

Let O be the center of the circle, let J and K be the points of tangency between circle O and DA and AB respectively, and let T be the point of tangency between circle O and EC. As all are radii of circle O, OJ = OK = OT = 1.

DA and AB are tangent to circle O at J and K, so ∠OJA = ∠AKO = 90°. As ∠JAK = 90° also, then ∠KOJ = 90° and AKOJ is a square.

EC is tangent to circle O at T, so ∠OTC = 90°. As ABCD and AKOJ are squares, both share vertex A, and AK and JA are coincident with AB and DA respectively, then O is colinear with AC. As AO, being the diagonal of a square, equals (1)√2 and AC = 4√2, OC = 3√2.

Triangle ∆OTC:
OT² + TC² = OC²
1³ + TC² = (3√2)²
TC² = 18 - 1 = 17
TC = √17

Let ET = x. As KE and ET are tangents from circle O intersecting at E, KE = ET = x.

Triangle ∆EBC:
EB² + BC² = CE²
(3-x)² + 4² = (√17+x)²
9 - 6x + x² + 16 = 17 + 2√17x + x²
2√17x + 6x = 25 - 17
(2√17+6)x = 8
x = 8/(2√17+6)
x = 8(2√17-6)/(2√17+6)(2√17-6)
x = 16(√17-3)/(68-36)
x = 16(√17-3)/32
x = (√17-3)/2

EB = 3 - (√17-3)/2
EB = (6-√17+3)/2
EB = (9-√17)/2

A = ((9-√17)/2)(4)/2 = 9 - √17 sq units

[glorytoturkiye]

such a good question and demonstrating

[FATFATHER]

tan(2theta)=4/(3-x) is good as well
as let angle NOE=theta

[nzm]

9:35 can you explain where did the "x" in esquire route of 17 come from please?🤔

[josephsalinas]

Eu fiquei em dúvida se a diagonal do quadrado passa no centro da circunferência.