Geometry Problem 1509. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Extend FB to point G such that BF=BG. Then we get CF=CG=CD, hence C is center of circle passing through points D,F and G. Since Angle DCF=90, Angle DGF=90/2=45.

ReplyDeleteHence Triangle ADG is right isosceles. We get AD=AG=AF+BF+BG=2+5+5=12.

Very elegant solution Pradyumna, much better than my own

DeleteThank you Sumith for your kind comment.

DeleteExtend CF to G

ReplyDeleteLet the side of the square be a, let GF = t and AG = q

From similar triangles BCF, AGF and GCD

5/a = 2/t = a/(q+x).....(1)

Since AFCD is concyclic, t(a+t) = q(q+x)....(2)

Substituting from (1) (2a/5)(a+2a/5) = (a^2/ 5 - x)(a^2/ 5)

14/5 = a^2/ 5 - x and a^2 = 14 + 5x...(3)

But x^2 +2^2 = 2.a^2

Therefore x^2 + 4 = 28 + 10x from (3)

x^2 - 10x - 24 = 0

(x+2)(x-12) = 0

So x = 12 since x = -2 is not permissible

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/tM61SHQ3HgmUEZrp9

ReplyDeleteLet H is the project of C over AD

Note that triangle FBC congruence to DHC ( case AA)

So HD=BF=5 and CH=AB=7

So AD=12

Good solution, Peter

DeleteThank Sumith

Deletec.t.e.o.https://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9

ReplyDeletehttps://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9

ReplyDeleteAnother Solution.

ReplyDeleteComplete Rectangle ADYX.

Then Triangle BCF, XFE and YED are congruent ASA.

So XE = DY = FB = 5 and hence, FX = EY = 7. Therefore AD = XE + EY = 5 + 7 = 12

Sumith Peiris

Moratuwa

Sri Lanka

Note that E is on XY

Delete