Japanese Math Olympiad Problem | A Very Nice Geometry Challenge

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال AD=1 DB=AD+c=1+c AB=d CD=1+c في المثلث ADC 1+c=2cos48 في المثلث ABD d^2=(2sin12)^2/2 d=2sin12 sinx=1/2 x=30

[@holyshit922]

We only need to play with measures of angles Sum of measures of angles in triangle is 180 Straight line angle measure is 180 At first sight i thought it would be easier It is still quite easy but more difficult than i thought at first sight

[@georgexomeritakis2793]

I solved this by using side BD and construct an isosceles triangle BED with angles 48,84,48 and point E looking downwards. Then observe that triangle ADE is equilateral while triangle AEB is isosceles.

[@kinno1837]

I like to see questions like this that are instantly clear after adding lines.

[@jejnsndn]

How to think like this?

[@MathBooster]

By practicing a lot of problems, you will be able to think faster.

[@mathisnotforthefaintofheart]

5:59 "suppose theta"....but you already used theta in the original angle ABD. You should not use the same letter for angles that you don't know they are equal (even though at the end that turns out to be true).

[@giuseppemalaguti435]

Risulta dalla figura AD=AC=a..AB=b...teorema dei seni a/sin(θ+72)=b/sin24...a/sinθ=b/sin12...divido le equazioni...sinθ/sin(θ+72)=sin12/sin24=1/2cos12... sviluppo i calcoli risulta ctgθ=(2cos12-cos72)/sin72=√3...θ=30

[@d4ig0]

como posso aprender a calcular dessa forma?

[@User-jr7vf]

I did it this way. However her way is much more beautiful and is the practical way if you are thinking about an exam.

[@akifbaysal9141]

Found it using trigonometry but not as clean as in pure geometric solution given here..

[@advaykumar9726]

What happens if we draw a circle through A B and D with C as center?

[@manojkantsamal4945]

42

[@robertlynch7520]

Here's the trigonometric solution ... #!/usr/bin/perl; # on a MAC; # include your favorite trig library if needed, such as 'use Math::Trig;' my $ab = 1; printf "ab = %f ", $ab; my $ah = $ab * cos deg2rad( 12 + 36 ); printf "ah = %f ", $ah; my $bh = $ab * sin deg2rad( 12 + 36 ); printf "bh = %f ", $bh; my $ac = 2 * $ah; printf "ac = %f ", $ac; my $ef = $ac * sin deg2rad 36; printf "ef = %f ", $ef; my $af = $ac * cos deg2rad 36; printf "af = %f ", $af; my $af_ah = abs( $af - $ah ); printf "|af-ah| = %f ", $af_ah; my $bh_ef = abs( $bh - $ef ); printf "|bh-ef| = %f ", $bh_ef; my $theta = rad2deg atan2( $bh_ef, $af_ah ); printf "theta = %f ", $theta; my $alpha = 180 + $theta - 48 - 84; printf "alpha = %f ", $alpha; my $beta = 180 - (96 + $alpha); printf "beta = %f ", $beta; ---------- OUTPUT ---------------------------------------------------------------------- ab = 1.000000