Japanese Math Olympiad Problem | A Very Nice Geometry Challenge

Japanese Math Olympiad Problem | A Very Nice Geometry Challenge
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[@MarieAnne.]

You pretty much solved it the way I did, except that I calculated ratio a/b slightly differently. I started out the same way up to finding the ratios: a/b = (a+b)/a But instead of cross-multiplying and finding a in terms of b, I calculated a/b directly as follows: a/b = 1 + b/a Multiply both sides by a/b: (a/b)^2 = (a/b) + 1 (a/b)^2 − (a/b) −1 = 0 We have a quadratic in a/b, so we can solve for a/b using quadratic formula: a/b = (−(−1) ± √(1−4(1)(−1))) / (2(1)) = (1 ± √5) / 2 Since a/b is the ratio of two positive numbers, it also must be positive: a/b = (1 + √5) / 2 Now we use law of sines in △ADC sin(18+x)/sin(18) = a/b = (1+√5)/2 sin(18+x) = (1+√5)/2 * sin(18) sin(18+x) = (1+√5)/2 * (√5−1)/4 = (5−1)/8 = 1/2 18+x = 30°, 150° *x = 12°, 132°* Both solutions work (I graphed them both in Desmos). Of course, in the second case angles ACB and ADB are acute, not obtuse as shown in diagram.

[@mnmn5580]

Excellent demonstration. I propose a variant after the time 6:34 where you demonstrate that a/b=φ (the golden ratio). This fact and the 18° angle made me think of the golden triangle, i.e. the isosceles triangle in which the angle included between the two equal sides is 36° (that is, two angles of 18°) and the ratio between each of the two equal sides and the third side is φ. Then draw the segment BM = BC such that ABM=18°. Thus we have a triangle CBM which is isosceles by construction and with the angle between the two equal sides CBM=36°. As stated above, the ratio BC:MC=BM:MC=φ. Knowing that a/b=φ also implies (a+b)/a=φ we can conclude that MC=a. By construction AB is perpendicular to MC and divides it into two equal segments. Said N the point of intersection between AB and MC we have NC=MC/2=a/2 and therefore AC=2NC for which the right triangle ANC is half of an equilateral triangle; therefore x+18°=30° and therefore x=12°.

[@marioalb9726]

a / (a+b) = b / a a² = b ( a+b) a² - ab - b² = 0 sin 18°/b = sin (18+x) / a sin (18-x) = a/ b sin18° 18° - x = asin ( 0,309 a/b ) x = 18°- asin ( 0,309 a/b ) I couldn't make more calculations Above we have the ratio between 'a' and 'b' and the formula of 'x' depending on 'a/b' Puting this formulas in an Excel, we obtain the angle 'x'

[@alexey.c]

What if x + 18° = 150°?

[@dave_9041]

that would make x greater than 90 degrees and u cant have 2 angles greater than 90 degrees in a triangle

[@alexey.c]

@@dave_9041 That means the other angles will be less than 90°, which is allowed by the conditions of the problem.

[@dave_9041]

@@alexey.c but angle ADB is greater than 90 degrees and if it were less than 90 degrees and x was greater than 90 then CD and AB would never intersect

[@alexey.c]

@@dave_9041 Nope. The solution tells you that both ADB and ACB are less than 90°. And CD and AB perfectly intersect in this case.

[@MarieAnne.]

@@alexey.c You're absolutely correct. I tried both solutions in Desmos, calculating all the sides using the appropriate ratios, then plotting each point and it worked for both x = 12 and x = 132. Of course, in the latter case, angles BDA and BCA are acute, and not obtuse as shown in diagram.

[@birandkoray]

i disliked the question, i have to calculate sine of something, is this the olympic level?

[@MathBooster]

Why cannot eliminate ? I don't undersand. If you are talking about general solutions of sin(x+18°) = 1/2. Then x+18° can be 30°, 150°, 390°, 510° ...... and so on. But from the given figure, x+18° = 30° will be only possible solution.

[@birandkoray]

@@MathBooster i mean sin18

[@MathBooster]

If you don't remember the value of sin18° [because 18° is not very common angle] , then you need to calculate it . It is easy to calculate the value of sin18° using trigonometric formulas.

[@じーちゃんねる-v4n]

36° and 72° are quasi-famous angles that appear in regular pentagons and often appear in math Olympiads. You can find the answer by using the formula cos36°cos72°=1/4 or cos36°-cos72°=1/2 without memorizing the numbers. If cos36°=c then 4c^3-3c=cos108°=-cos72°=-2c^2+1 ∴4c^3+2c^2-3c-1=(c+1)(4c^2-2c -1)=0, 4c^2-2c-1=0 ∴cos72°=2c^2-1=c-1/2

[@padraiggluck2980]

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