Poland Math Olympiad | A Very Nice Geometry Problem

Poland Math Olympiad | A Very Nice Geometry Problem
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[@hongningsuen1348]

For quick but less precise solution: 1. Triangles ABC and CBE are similar (AAA) Hence AB/BC = BC/BE Hence BC = sqrt (AB.BE) = sqrt (5 x3) = sqrt (15) 2. Use sine rule to find angle ACB of triangle ABC sin(ACB)/AB = sin45/BC sin(ACB) = [sqrt(2)/2 x 5]/sqrt (15) = sqrt (5/6) = 0.91287 Hence angle ACB = arcsin 0.9128 = 65.91 or (180 - 65.9) = 114.09 As angle ACB > 90, 65.91 discarded. 3. In triangle ABC, angle (ABC) = 180 - 45 - 114.09 = 20.91 (angle sum of triangle) 4. Side of square = BE x sin(ABC) = 3 x sin20.91 = 1.071 5. Area of square = 1.071^2 = 1.147 (Your answer is around 1.146)

[@sinsn2147]

For 45 deg is half of EDC, if a circle centered on D with radius ED=x, A must lies on the circle. Thus, AD is also at length of x, which means the height of ∆AED with base of AE will separate AE equally. Let AE and it's height meet at point G, EG=1 ∆GED is similar to ∆FBE, it has 3/BF=x/1 => BF=3/x Looking into ∆BEF, we have 9-x^2=9/x^2 Let x^2=S, which is the answer itself S^2-9S+9=0 S=(9+√ 45)/2 or (9-√45)/2 If D must inside ∆ABC (I don't see why), apparently the former answer is larger than 2*2=4. Thus, the area is S=(9-√45)/2

[@b07_ashishsuradkar49]

What a magnificent problem.Thank you sir.

[@timc5768]

Once again, I missed your more elegant sol'n. However, by elementary trig., I think you could say: Let angle ABC = u, Then X = 3sin(u) , Area = 9{sin(u)]^2 (eq'n 1) In triangle AEC, EC = 3[sqrt(2)]sin(u), ( Pythag.), angle AEC = 45 + u, so angle ECA = 90 - u , (sum of angles in triangle), (3[sqrt(2)]sin(u))/sin(45) = 2/[sin(90 - u], (sin rule), so 6sin(u) = 2/cos(u), sin(u) cos(u) = 1/3, [{sin(u)]^2}][{cos(u)}^2] = [sin(u)]^2 - [sin(u)]^4 = 1/9, and let 'p = [sin(u)]^2', (eq'n(2) so p^2 - p = - 1/9, [p - 1/2]^2 = 1/4 - 1/9 = 5/36, p = 1/2 + or - [sqrt(5)]/6, eq'n(3) , so from Eq'n (1,2,3), (and dismissing the larger root) Area = 9p = 9/2 - [3sqrt(5)]/2 = (1/2)[9 -3sqrt(5)]

[@zdrastvutye]

the path to the solution is that the y coordinate of point e must be the distance from point c: 10 print "mathbooster-poland math olympiad":l1=3:l2=2:sw=l1/(l1+l2) 20 dim x(4,2),y(4,2):w1=45:w2=sw:xb=0:yb=0:n=sqr(l1^2+l2^2):w2=70+sw:goto 70 30 w3=180-w1-w2:lbc=(l1+l2)*sin(rad(w1))/sin(rad(w2)):xc=lbc:yc=0 40 xa=(l1+l2)*cos(rad(w3)):ya=(l1+l2)*sin(rad(w3)) 50 xe=(xa-xb)*l1/(l1+l2):xe=xe+xb:ye=(ya-yb)*l1/(l1+l2):ye=ye+yb 60 dgu1=l1*cos(rad(w3))/n:dgu2=(ye-lbc)/n:dg=dgu1+dgu2:return 70 gosub 30 80 dg1=dg:w21=w2:w2=w2+sw:w22=w2:gosub 30:if dg1*dg>0 then 80 90 w2=(w21+w22)/2:gosub 30:if dg1*dg>0 then w21=w2 else w22=w2 100 if abs(dg)>1E-10 then 90 110 print w2:agesu=ye^2:print"die flaeche="; agesu:masx=1200/xa:masy=900/ya 120 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

[@user-nr6tt5he7g]

It's not clear for me, where you got DE

[@mireyajones810]

If I make a circle around the square, then AE and AC are tangents to the circle, therefore AE = AC = 2. Thus CE is found via Law of Cosine, CE^2 = 2^2 + 2^2 - 4Cos(45). CE^2 = 8 - 2*root 2; CE = X * root 2. Therefore CE is approximately 1.3; NOT (root 15 minus root 3)/2. Your thoughts?

[@user-nr6tt5he7g]

Why do you think they are tangent? It just isn't true.

[@hongningsuen1348]

AC and AE can be tangents. But it will lead to a different BC as derived from similar triangle corresponding side proportionality equation, i.e. sqrt 15 hence BC^2 = 15. By cosine rule, your BC^2 = 5^2 + 2^2 - 2(5)(2)cos45 = 29 - 10sqrt2 = 14.857864

[@user-uc2qy1ff2z]

a=3 b=2 c is side of square u= ABC c = a sin(u) h= c*(a+b)/a= (a+b) sin(u) tg(u)=h/(a cos(u) +c + h tg(п/4-u)) Because sharp angle of right-angled triangle, which is needed to be added to ABC to RAT is 90-45-u tg(u)(a cos(u)+a sin(u)+(a+b)sin(u)(1-tg(u))/(1+tg(u))) = (a+b) sin(u) (a(1+tg(u))+(a+b)tg(u)(1-tg(u))/(1+tg(u))) = (a+b) x = tg(u) a(1+x)²+(a+b)(x-x²)=(a+b)(1+x) a+2ax+ax²+ax-ax²+bx-bx²=a+b+ax+bx 2ax-bx²=b bx²-2ax+b=0 b(tg²(u)+1)=2a tg(u) b/cos²(u)= 2a tg(u) b/a= 2sin(u)cos(u) b²/a²=4sin²(u)(1-sin²(u)) sin⁴(u)-sin²(u)+b²/(4a²)=0 sin²(u)=(1±sqrt(1-b²/a²))/2 sin²(u)=(1±sqrt(a²-b²)/a)/2 =(1+sqrt(5)/3)/2>1, which is impossible. S=c²=a²sin²(u)=(a²-a sqrt(a²-b²))/2 = (9-3 sqrt(5))/2