Find the shaded area in the quarter circle | Poland Math Olympiad Geometry Problem

[@marioalb9726]

Sine rule: sin α/√3 = sin 120°/R sin α = 1/2 ; α = 30°=π/6 Shaded area: A = ½αR² - ½R.½R.tanα A = ½.π/6.3² - ¼3²/√3 A = 2,3562 - 1,299 = 1,057 cm²

[@oscarcastaneda5310]

Awesome ! Law of Sines reveals the sector angle to be 30 degrees and from this we're able to find the sector area and the triangle area. Subtracting gives the desired answer. We Think Alike : )

[@ناصريناصر-س4ب]

The line PQ intersects the entire circle at point E. The triangle OEQ is right because OE=OQ*√3. Hence, the inscribed angle AEP measures 30°. Thus,

[@jimlocke9320]

ΔOEQ is a right triangle, but I do not see that you have proven that it is. We note that OE has length 3 by being a radius, OQ has a given length of √3 and

[@ناصريناصر-س4ب]

​@@jimlocke9320The line PQ intersects the circle at point E. It is not necessary that it be diagonally opposite point A. Let us assume that E' is the point diagonally opposite point A. Then the triangle E'OQ has side lengths of 3,√3,2√3. Hence

[@changryu8128]

You are wrong!

[@giuseppemalaguti435]

Ablue=Asett-Atriang=9π(60-arcsin1/2)/360-3√3/4=9π/12-3√3/4=(3/4)(π-√3)

[@zawatsky]

Достраиваем до прямоугольного дельтоида AOQP, который составлен из подобных 📐1;2;√3;30;60;90. Из него легко доказать, что ∠AOP=60°. Из т. Р на ОВ опускаем высоту РЕ, получаем 📐EPQ и 📐ОЕP, того же типа. ЕР=3/2, QP=(3/2)÷√3=√3/2,✔ QP=√3. Зная площади этих двух, вычислим площадь ▲OPQ, которую предстоит вычесть из сегмента. S(OPQ)=S(OEP)-S(EPQ). Площади подобных фигур относятся как квадраты соответствующих сторон, т. е., в нашем случае, как меньший катет EQ к большему ЕР или √3/2÷(3/2)=2√3/(2*3)=√3/3=1/√3. Имеем S(OPQ)(1-1/√3). Вычисляем базовую площадь: S(OPQ)=(√3+√3/2)(3/2)/2=1½√3*¾=3√3*3/(2*4)=9√3/8. 9√3/8÷√3=9/8. S(OPQ)=9(1-1/√3)/8. Теперь сегмент. Это 1/12 круга, т. е. πr²/12=9π/12=3π/4. Тогда искомая площадь S=¾π-⁹/₈(1-1/√3)=(6π-9(1-1/√3))/8.

[@AmirgabYT2185]

S=3(π-√3)/4=¾(π-√3)≈1,056

[@jimlocke9320]

Drop a perpendicular from P to OB and label the intersection as point R. Note that ΔOPR is a 30°-60°-90° right triangle. Let short side QR have length x. The long side of a 30°-60°-90° right triangle is √3 times as long as the short side, so PR has length x√3. Consider right ΔOPR. The hypotenuse is the circle's radius, 3. Side OR has length x + √3 and side PR length x√3. Apply the Pythagorean theorem and solve the resulting quadratic equation, finding that the positive root is x = (√3)/2. So, OR has length (3/2)√3 and PR has length 3/2. From the ratios of side lengths, we find that ΔOPR is also a 30°-60°-90° right triangle, with

[@kateknowles8055]

Thank you for another brain teaser. Extending PQ and AO to meet at A' , A'OQ is 90 degrees , OQA' = 60 degrees It is vertically opposite PQB. and QA'O= 30 degrees It is 180- 90 - 60 degrees in triangle A'OQ. sqrt (3) = tan( 60 degrees) so OA' = 3 AA' is a diameter AA' = 3+3 = 6 In triangle BPQ, P is on the circumference subtended by the arc BA' which measures 1/4 of the circumfence . Angle BPQ = (1/2) angle A'OB BPQ is 45 degrees. QBP = 180 - 60 - 45 = 75 degrees. ( Everthing is known, except for all that is necessary for a solution. Oh dear!) QA'O is 30 degrees. sine (30) = 1/2 = OQ / QA' so QA' = 2sqrt(3) =riangle APO is equilateral : AOP= 60 degrees angle OAP = angle OPA The sector is 1/6th part of pi . r^2 ( 1/6).3.3.pi = 3 pi /2 The sector POB is 3 pi /4 90 - 60 = 30 so In triangle POQ angle O is 30 degrees, OP is a radius =3 OQ = sqrt(3) area [POQ] = (1/2) OP.OQ.sin(30) = 3 sqrt(3)/4 Subtracting areas : The red shaded area = [3 pi /4] - [3sqrt(3)/4] = (3/4) . (pi - sqrt(3))