Find the length X | Math Olympiad Geometry Problem | Important Geometry and Algebra Skills Explained

[@EnnioPiovesan]

Triangle AEO is equilateral therefore AE=AO=EO=30. Let H be the projection of E onto the diameter AB. We have: AH=AO/2=15. EH=AH√3=15√3. HD=AD-AH=40-15=25. Applying the Pythagorean theorem to the triangle EHD we have: ED²=EH²+ED² --> ED²=3·15²+25²=1300 --> ED=10√13. Triangles EHD and POD are similar because EDH is in common and EHD=POD=90° (obvious reason of symmetry). Therefore: PD:ED=OD:HD --> x:10√13 = 10:25 --> 25x=100√13 --> x=4√13.

[@d-8664]

Use the cos rule to find length ED to be sqrt(1300). Then calculate angle EDO via the sine rule to be sin(Beta) = 15*sqrt(3)/sqrt(1300). Then x =10/cos(Beta) = 4*sqrt(13).

[@jimlocke9320]

Draw ΔOEM early on and prove

[@ΕυθύμιοςΠαπασταυρίδης]

EF ist gleich dem Kreisradius (als Seite eines einbeschriebenen 6-Ecks), also EF=30. Weiterhin ist EF // CD (Eigenschaft der einbeschriebeben 6-Ecke). Folglich sind die Dreiecke EPF und CDP zueinander aehnlich (3:2). Demnach ist DP:PE=2:3, also DP:ED=2:5. ED laesst sich aus dem Kosinoussatz im Dreieck EAD berechnen, da der Winkel EAD 60 Grad betraegt. Man bekommt ED=10 * Wurzel 13. DP=x=0,4*10 Wurzel 13=4 Wurzel 13.

[@sarantis40kalaitzis48]

Excellent.

[@ROCCOANDROXY]

In general, letting AC = CD = DB = a implies x = sqrt(13)/5 * a. Also, letting A be the area bounded by arc EF and the two intersecting lines we obtain A = 9/4 *(pi/6 - sqrt(3)/10) * a^2. Also, Area(Triangle(CPD)) = 3 * sqrt(3)/20 * a^2 implies Area(Triangle(CPD))/A = sqrt(3)/(15 * (pi/6 - sqrt(3)/10)).

[@denismilic1878]

the semi circle is divided into three ideal parts, angle is 60° points B and F with circle center make an equilateral triangle vith height 30√3/2 = 15√3 C(-10,0), F(15, 15√3), P(0, y) y = 10 * 15√3 / 25 y = 6√3 -> P(0, 6√3) x² = (6√3)² + 10² x = √208 = 4√13

[@vierinkivi]

Piirretään suorakulmainen kolmio, hypotenusa ED suorakulma halkaisijalla pisyteessä G. kateetti EG 15*sqrt3, kateetti GD 25. Hypotenusaksi saadaan sqrt(25^2+(15sqrt3)) = 10sqrt13 2/5*10sqrt13 = 4sqrt13

[@marioalb9726]

Angle EOA = Angle FOB = 180°/3 = 60° R = 20/2+20 = 30 cm EF = R = 30 cm Pythagorean theorem: ED²=(30/2+20/2)²+(30cos30°)² ED² = 25²+25,98² ED = 36,0555 cm = CF Similarly of triangles x/10 = 36,0555/25 x = 14,42 cm ( Solved √ )

[@sonam7choi]

삼각형 CDE 에 코사인법칙을 쓰면 ED=10ROOT(13) PX=ED*2/5=4ROOT(13)

[@AttilioGrassotti]

Another solution. Call M the projection of E on AB. ME = 15. sqrt(3). Angle EDM = angle PDO = arctan(15.(sqrt(3)/25) = arctan(3.sqrt(3)/5). DP = 10/cos(arctan(3.sqrt(3)/5) ) = 14,4222051 = 4.sqrt(13).

[@Latronibus]

I see now why you set the semicircle to be the size that it is: if the center is (0,0) then you're getting the distance between (0,rsqrt(3)/5) and (r/3,0), so the numbers work out more cleanly if r is a multiple of 5 and a multiple of 3. Having r be 15 would have been more or less equally as clean though.

[@АндрейБерзенин]

ED=✓(EO²+OD²-EO•OD•cos120°)

[@n.662]

Magnificently! Very nice solution! 🎉

[@MathBooster]

Thank you 🙂

[@El_Girasol_Fachero]

New subscriber 🌟 greetings!!👋🏻

[@MathBooster]

Thank you 🙂

[@mr.gamerboss6398]

3^n+2^n=793 then x=? Plz solve this problem

[@sagnikbiswas3268]

Similar triangles are helpful

[@buziga236]

Nice

[@kaliprasadguru1921]

Sir , whereas you are explaining in great detail, you should have told a line as to why angles EFC and FCD will be equal for clarity of average students .

[@samimarzou]

I agree. He's assuming AB and EF are parallels. I know they are but I was stuck there

[@MathBooster]

The figure is symmetric so EF should be parallel to AB. Otherwise you can easily see that angle AOE is equal to angle OEF that is 60 degree, that's why EF should be parallel to AB.

[@kaliprasadguru1921]

Very small things you are going in detail . This main issue should have been explained . Symmetry is not a good explanation . Parallel due to equal angles of 60° holds good for correct analytical reasoning .This is a main issue and you skipped simply saying that angle EFC is equal to angle FCD . This reminds me of my matriculation days of the early 1960's I thank you for that .Namaste .

[@saurabhsrivastav2567]

The written comments interfere on screen video

[@MathBooster]

You can turn them off by clicking the cc button in the top right corner.

[@saurabhsrivastav2567]

@@MathBooster what I meant was the transcript of voiceover on the KZbin screen doesn't let see the working of maths problem

[@elmurazbsirov7617]

Bakıdan salamlar.Əla həll etdiniz.Çox sağ olun.