Chinese Math Olympiad 1991 Problem | Geometry Question | Important Geometry Skills Explained

Chinese Math Olympiad 1991 Problem | Geometry Question | Important Geometry Skills Explained
Join this channel to get access to perks:
/ @mathbooster
MEMBERS OF THIS CHANNEL
••••••••••••••••••••••••••••••••••••••••
Mr. Gnome
Sambasivam Sathyamoorthy
ஆத்தங்கரை சண்முகம்
Εκπαιδευτήρια Καντά
AR Knowledge
堤修一
Sidnei Medeiros Vicente
Mark Ludington
Saunak Swar

[@davidellis1929]

SSA is not sufficient to prove ACE and ABE are congruent. But the Angle Bisector theorem guarantees AC:AB=CE:EB=5z:5z=1:1. You get a similar and slightly simpler solution by drawing EG || BD instead of DF || AE. An entirely different approach uses Mass Points. Start with weights B=3 and D=5. The puzzle pieces fit together if we assign A=2 and E=6, yielding C=3, establishing CE=EB quickly. The Angle Bisector theorem is the final step, making ABC isosceles.

[@jeanmarcbonici9525]

Very interesting approach

[@leomourao75]

It is possible to find the areas of the triangles CDH = 6S and HCE = 5S considering the following areas: DEH = 3S ADH = 9S AHB = 15S BBB = 5S Use the proportional lines (3x/x and 3y/5y) to find these areas. Then areas ACE = AEB = 20S.

[@DB-lg5sq]

شكرا لكم يمكن استعمال L مسقط E على(DC) بتواز مع(AE)وبعد تطبيق طاليس نجد L منتصف [CF]وبالتاليE منتصف[BC].....

[@MarieAnne.]

For my solution, I used angle bisector theorem and law of cosines. After determining that AH = 2x, HE = x, BH = 5y, HD = 3y, I used angle bisector theorem which states that an angle bisector divides opposite side of triangle in same ratio as the other two sides of triangle. So in △ABD, we get: AB/AD = BH/DH = (5y)/(3y) = 5/3 → AB = 5k, AD = 3k Let ∠AHD = α. Then ∠AHB = 180−α Using law of cosines in △ADH, we get: cos α = (9x² + 9y² - 9k²) / (2(3x)(3y)) = (x² + y² − k²) / (2xy) ........ (1) Using law of cosines in △ABH, we get: cos(180−α) = (9x² + 25y² - 25k²) / (2(3x)(5y)) = (9x² + 25y² − 25k²) / (30xy) Use trig identity: cos(180−α) = −cos α (9x² + 25y² − 25k²) / (30xy) = (k² − y² − x²) / (2xy) 9x² + 25y² − 25k² = 15 (k² − y² − x²) y² − k² = −3x²/5 Plugging this back into equation (1) we get: cos α = (x² − 3x²/5) / (2xy) = x/(5y) In △BEH, we have ∠BHE = ∠AHD = α (vertical angles are congruent), and the two sides adjacent to α are HE = x, BH = 5y But we know that cos α = x/(5y) = HE/BH. So, △BEH must be a right triangle, with HE adjacent to angle α , and with hypotenuse BH. Therefore, ∠HEB = 90° → ∠AEC = ∠HEC = 180°−90° = 90° In △ACE, ∠AEC = 90° and ∠EAC = 70°/2 = 35° ∠C = 180° − 90° − 35° = 55°

[@jeanmarcbonici9525]

Nice thank you

[@sivajirangisetty5103]

Tq sir i learnt so much concepts with your channel

[@Abby-hi4sf]

Thank you for your time

[@derwolf7810]

The triangles AEC and AEB are congruent if and only if the 35° angle opposes the longer side (AE < EC). For example in case your drawing would be accurately representing the real situation (in the sense that EC < AE < AC), then the circle around E with radius r = EC intersects side AC twice (in C and C'), resulting in two clearly non-congruent triangles AEC and AEC', with shared angle A=35°, shared size AE and AC = r = AC'. Edit: Corrected a flaw.

[@СтасМ-ъ8б]

Очень красивое и элегантное решение!

[@kaliprasadguru1921]

Sir , how can ∆s ACE & ABE be congruent ? . It is ok ,CE=EB and AE is common side . But angles CAE and BAE are not the included angles . It is not conforming to S - A - S condition. Hence apply theorem of angle bisector dividing the sides in propertion to......i.e. AB/AC =BE/CE . But BE/CE =1 .So AB/AC = 1 i.e. AB=AC . Hence isosceles.

[@MathBooster]

SSA works in some special cases (not always). Also your method is very perfect.

[@aamethod7079]

Hello sir e ke power y ka differentiation kya hoga

[@임석근-t5j]

4엑스는 8와이 1엑스는 5와이....180 나누기 6는 30 .... 150 나누기 2는 75입니다.

[@alinayfeh4961]

I appreciate solution

[@krishnamoyghosh6047]

The last step is wrong. If two sides of and included angle of an triangle are equal to corresponding two sides and included angle then triangles are congruent.

[@MathBooster]

SSA also works in some special cases (not always). So, you can do the last step by using the concept 'angle bisector is also bisecting the third side' so the triangle will be isosceles.

[@kinno1837]

10:48 Why there are 35 degrees?

[@MathBooster]

Because angle A is 70° and AE is angle bisector (angle bisector bisects the angle in equal parts). So, those two angles will be equal and they will be 70°/2 = 35°