Math Olympiad | Find the area of semicircle inside a square | 2 Different Methods

Math Olympiad | Find the area of semicircle inside a square | 2 Different Methods
MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : / @calculusbooster
Math Hunter : / @mathshunter
--------------------------------------------------------------------------------
Join the channel to become a member
/ @mathbooster

[@sena23]

3rd and the easiest solution: Draw |OQ| line, which is the radius r of the semi circle. Since the angles APD, PAB, QOC are equal, AQOB and OQPC are similar deltoids. [PC] can be calculated using similarity: [AB] / [OC] = [QO] / [PC]. So, [PC] = r.r/2.r => [PC]=r/2 Area of triangle=|AD|.|DP|.1/2 6 = 2r.(2r-r/2).1/2 6 = r.3r/2 4 = r² semicircle area = 4.π/2 = 2π old but gold here ;)

[@Grok-vg9qg]

Nice!

[@Irtsak]

Let α the side of the square ABCD => α=2R and DP=x. Obviously PC=α-x AB=AQ=α ( tangent segments from A) and PQ=PC=α-x (tangent segments from P) AP=AQ+QP=α+α-x=> *AP=2α-x* Area of triangle ADP =6 => 1/2 AD⋅DP=6=> 1/2 α⋅x=6 => *αx=12* (1) Pythagoras theorem in orthogonal triangle ADP => AD²+DP²=AP²⇒α²+x²=(2α-x)² α²+x²=4α²-4αx+x² 3α²-4αx=0 3α²-4⋅12=0 cause (1) So α=4 and R=α/2=2 Area of the semicircle = (πR²)/2=(π⋅2²)/2=2π

[@michaeldoerr5810]

I now have a question: wouldn't the second method be simplified by the two tangent theorem? I came across a PreMath video that tackles that similar problem.

[@juanalfaro7522]

Let L=square sides, a-PC. Then L=2R -> R=L/2. We have AB=AQ=L and PQ=PC=a. Then [L-a] ^2 + L^2 = [L+ a] ^2 -> (L^2 - 2aL + a^2) + L^2 = (L^2 + 2aL + a^2) -> L^2 = 4aL -> L=4a, a=L/4. Now L*(L-a)/2 = 6 -> L*(L-a) = 12 = L * 3L/4 -> L^2 = 16 -> L=4 -> R=2 --> Area (semicircle) = Pi * 2^2/2 = 2*Pi = 6.28 sq. units (39.3% of square).

[@jimlocke9320]

At 7:25, we have 8Rx = 4R². We can safely assume that R is not 0 and factor out R from both sides, leaving 8x = 4R and x = R/2. Now we return to 6 = 2R² - xR and substitute R/2 for x, leaving 6 = 2R² - (R/2)R = 2R² - R²/2 = 3R²/2, (6)(2) = 3R², R² = 4. From there, proceed to calculate the area of the semicircle as (1/2)πR² = (1/2)π(4) = 2π. An interesting observation can be made about right ΔADP. The lengths of the sides are AD = 4, DP = 3 and hypotenuse AP = 5. So, its sides form the Pythagorean triple 3-4-5. If the area of right ΔADP is changed, the new triangle will be similar to a 3-4-5 right triangle. The ratio of area of the ΔADP to the semicircle will remain 6:2π = 3:π

[@alegoncalves472]

Wow!!! Thanks for noticing 😊😊😊😊

[@josephsalinas6725]

Esse segundo método foi muito bom !

[@prossvay8744]

Area of the semicircle=2π. Thanks ❤.

[@mathiq56]

I love learn math with you

[@murdock5537]

φ = 30°; ∎ABCD → AB = BC = BO + CO = a/2 + a/2 = r + r = CD = CP + DP = AD = a AP = AQ + PQ; sin⁡(OBA) = sin⁡(AQO) = sin(POA) = sin(OQP) = sin⁡(3φ) = 1; area ∆ ADP = 6 BAO = OAQ = COP = POQ = ϑ → sin⁡(ϑ) = √5/5 → cos⁡(ϑ) = 2√5/5 → tan⁡(ϑ) = 1/2 → CP = a/4 → DP = 3a/4 → (1/2)(3a/4)a = 6 → a = 4 → a/2 = 2 = r → area semi circle = 2π

[@JohnPeter-j1s]

I love math ❤❤

[@GetMeThere1]

LOL. I'm embarrassed that it took me almost the entire video to see that PQ = PC, even though I knew easily that AB = AQ. I just couldn't see it!

[@quigonkenny]

Same here.

[@comdo777]

asnwer=35cm

[@michaeldoerr5810]

Golly the first method seems to be a lot easier