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Let ABC be a triangle with angle B = 90 degrees. Let D, E be points on AB such that BD = BC and AE = AC (they lie in the order B-E-D-A). Let F, G be the feet of perpendicular from E and D onto BC and AB respectively. If EF = 12 And DG = 11, what is DE?

While there are a ton of similar triangles, using their ratios did not lead to the nicest of equations (at least the way I did it). Given that this is a problem from a regional level olympiad with strict time controls (75 minutes for 8 questions) there is probably a clever way to utilize AC=AE and BC=BD which i seem to be missing

Wafi Marzouq Mohammad
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I'll assume, for the text to be meaningful, the right angle to be at C, and G to be the foot of the perpendicular from D to AC.

Set $AC=a$, $BC=b$, $AB=x=\sqrt{a^2+b^2}$. From the similarity of triangles $ABC$, $ADG$, $EBF$ we get: $$ (x-a):12=x:a, \quad (x-b):11=x:b, $$ that is: $$ a^2=(a-12)x, \quad b^2=(b-11)x. $$ Adding these two equations we get: $$ a^2+b^2=(a+b-23)x=x^2. $$ whence: $$ a+b-23=x $$ and finally: $$ DE=a+b-x=23. $$

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Intelligenti pauca
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