Poland Math Olympiad | A Very Nice Geometry Problem

Poland Math Olympiad | A Very Nice Geometry Problem

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[timc]

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)]

[NickDos-rf]

It's not clear for me, where you got DE<AE from. 😞 (at 10:07 )
Actually it is not true, unless you assume that the whole square is inside the angle BAC!
With the additional assumption I'd argue as follows.
Consider the halfline with the orgin A containing AC and imagine yourself a square with diagonal EC. If we slide C along the halfline, the vertex D will lie inside the angle BAC only for C in some interval. The end points of the interval lead to a square in which D lies on EB (gives the shortest diagonal) or on the halfline (gives the longest diagonal). So, the diagonal's length is between sqrt(2) and 2, whence the square's side is between 1 and sqrt(2). Only the second value belongs to [1, sqrt(2)].

[b_ashishsuradkar]

What a magnificent problem.Thank you sir.

[hongningsuen]

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)

[sinsn]

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

[mireyajones]

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?

[shaozheang]

draw a circle around ACDE. AC should be 2.

[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
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
40
50
60
70 gosub 30
80 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=";
120 if masx<masy then mass=masx else mass=masy
130 x(0, 0)=0:y(0, 0)=0:x(0, 1)=lbc-ye:y(0, 1)=0:x(0, 2)=xe:y(0, 2)=ye
140 x(1, 0)=lbc:y(1, 0)=0:x(1, 1)=lbc-ye:y(1, 1)=0:x(1, 2)=x(1, 1):y(1, 2)=ye
150 x(2, 0)=lbc:y(2, 0)=0:x(2, 1)=x(2, 0):y(2, 1)=ye:x(2, 2)=x(2, 1)-ye:y(2, 2)=ye
160 x(3, 0)=lbc:y(3, 0)=ye:x(3, 1)=x(3, 0):y(3, 1)=0:x(3, 2)=xa:y(3, 2)=ya
170 x(4, 0)=xe:y(4, 0)=ye:x(4, 1)=x(4, 0)+ye:y(4, 1)=ye:x(4, 2)=xa:y(4, 2)=ya
180 goto 200
190 xbu=x*mass:ybu=y*mass:return
200 for a=0 to 4:gcol a:x=x(a, 0):y=y(a, 0):gosub 190:xba=xbu:yba=ybu
210 for b=1 to 3:ib=b:if ib=3 then ib=0
220 x=x(a, ib):y=y(a, ib):gosub 190:xbn=xbu:ybn=ybu:goto 240
230 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
240 gosub 230:next b:next a:print 190
250 r=ye/sqr(2):gcol 8:circle xbu, ybu, r*mass
260
mathbooster-poland math olympiad
114.094843
die flaeche=1.14589803
3.87298335
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[a]

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 = (a+b) sin(u)
= (a+b)
x = tg(u)


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