Math Olympiad | A Very Nice Geometry Problem

Poland Math Olympiad | A Very Nice Geometry Problem
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[@florianbuerzle2703]

Solved it in my head 😂 The height is 4 by the Pythagorean theorem, then call the side of the square x and by similarity 3/4 = 0.5x / (4 - x). Solve for x gives x = 12/5, so x²=144/25 😊

[@RahulKumar-id5cq]

Can also be solved in following way:

[@Irtsak]

Math’s Booster solution is very good and explanatory

[@quigonkenny]

Let s be the side length of the square DEFG. As CA = AB = 5, ∆ABC is an isosceles triangle. Draw AH, such that H is the point on BC where AH is perpendicular to BC. Let J be the point where AH and ED intersect. As BC is the base of the isosceles triangle and A is the opposite vertex, by rule AH bisects ∆ABC, creating two new congruent right triangles ∆BHA and ∆AHC.

[@prossvay8744]

Area of the triangleABC

[@RahulKumar-id5cq]

We could also solved in this way:

[@alexbayan8302]

This problem is easy enough to be a slightly hard SAT problem.

[@mauriziofenderico8348]

Really like the step by step explanation and the graphics. Well done?😃😄

[@AmirgabYT2185]

S=5,76

[@michaeldoerr5810]

Honestly that is an explanation that makes more sense than all of the comments so far. All that had to be done was to bisect both the triangle and square and you have an AA similarity. This is resolved by Pythagorean Theorem and this is before similar sides are compared. I hope that I got this right. I really wished that I knew of this in high school. Please tell me if that makes sense.

[@victoroliveira3705]

X= 2.4

[@johngutwirth7706]

That was funny !