Russian Math Olympiad Problem | A Very Nice Geometry Challenge

[@kinno1837]

We can use sine law and cosine law to find theta. Angle BDA=180°-36°-18°=126°(angle sum of ∆) sin126°/1=sin36°/AD AD=sin36°/sin126° Angle ADC=36°+18°=54°(exterior angle of ∆) cos54°=(AD²+5-AC²)/2•AD•√5 AC≈1.902113033 sin54°/AC=sinθ/AD θ=18°

[@jimlocke9320]

Drop a perpendicular from A to BC and label the intersection as point F. Consider right ΔABF. Length AF = (AB)(sin(36°)) = (1)(sin(36°)) = sin(36°).

[@georgexomeritakis2793]

You don't need to draw and auxillary line here. We know from previous puzzles that sin 18 = ((sqrt(5)-1)/4 and sin 54 = (sqrt(5)+1)/4 so we can calculate the ratio AB/BD. Then calculate the ratio BC/AB and prove that triangles ABD and CBA are similar.

[@michaeldoerr5810]

Just one question, does your method make use of the Law of Sines? I could be wrong.

[@User-jr7vf]

Wow this is the first time I see Math Booster doing this. Would it be valid?

[@femalesworld2]

Why phita=18°?

[@kinno1837]

He just used an unknown to find the length of DE,not the answer of the question.

[@Cricketdoctor_1999]

18

[@prime423]

A good Mathlete would know the Sine18=sqr5 -1/4.After that, the problem is trivial. The formula is derived from a 72-72-36 triangle. That ,in fact has to do with a polygon. Which one?

[@jimlocke9320]

Let's try a regular pentagon. A side is (√5 - 1)/2 times as long as a diagonal (a diagonal connects 2 non-adjacent vertices). You can construct a 18°-72°-90° triangle within a regular pentagon with a diagonal as the hypotenuse and a half side opposite the 18° angle. The half side is (1/2)(√5 - 1)/2 = (√5 - 1)/4 times as long as the diagonal, so sin(18°) = (√5 - 1)/4.

[@comdo777]

asnwer=16cm isit

[@comdo777]

sam why asnwer=18cm hmm gmm

[@CapnbloodBeard0]

18

[@michaeldoerr5810]

Im just wondering: could you actually use the exterior angle theorem? Or is that redundant due to the triangle being a special triangle constructed by a pentagon?