Romania Math Olympiad Problem | A Very Nice Geometry Challenge

Romania Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Methods

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Romania Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Methods
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Пікірлер: 21

@jimlocke9320 6 ай бұрын

At 7:20, drop a perpendicular from D to AC and label the intersection as point F. Note that ΔCDF and ΔCBE are similar. Corresponding sides CD and CB are in the ratio 4:6 = 2:3. Corresponding sides DF and BE must be in the same ratio, so DF = (2/3)(a√3)/2 = (a√3)/3. ΔADF is a special 30°-60°-90° right triangle, with X as the hypotenuse. The hypotenuse is 2/√3 times as long as the longer side. So, AD = X = (2/√3)(a√3/3) = 2a/3. We have already found that a = 6/√7, so X = (2)(6/√7)/3 = 4/√7, as Math Booster also found.

@Mediterranean81 3 ай бұрын

Area of ABC = sqrt 3 a^2/2 Area of ABD= sqrt 3*ax/4 Area of ADC = sqrt 3*ax/2 sqrt 3 a^2/2 = sqrt 3*ax/4 + sqrt 3*ax/2 a^2/2=ax/4+ax/2 2a^2=ax+2ax 2a^2=3ax 2a=3x a=3x/2 law of cosines in ABD 4=a^2+x^2-2ax cos 60º 4= (3x/2)^2+x^2-2(3x/2)x*1/2 (4= 9x^2/4 +x^2-3x^2/2)×4 16=9x^2+4x^2-6x^2 16=7x^2 16/7=x^2 x=4/sqrt 7

@Mediterranean81 3 ай бұрын

@murvetaykac7041 4 ай бұрын

We can find this.In ABC cosinus theorem and 6^2=a^2+(2a)^2_2 .a.2a.cos120 .a^2=36/7 .Than a×2a_2.4=x^2 .Put a^2 in this equality.x^2=26/7.

@petrusneacsu9858 6 ай бұрын

Or another method: Extend AB to point E with BE=AB=a. The triangle AEC is isosceles with two angles of 30 degrees and one angle of 120 degrees. Extend AD until it intersects EC at point H. AH is the bisector, median and height in the isosceles triangle AEC. It follows that D is the intersection of the medians in AEC, that is, it is the center of gravity of the triangle AEC. It immediately follows that DC=2*BD=4. Now we focus on the triangle AHC. We know that AH=a because the 30 degree angle theorem says that AH=AC/2 because the angle ACH=30 degrees. I denote x/2 by y to make calculations easier. So AH=3*y=a. So AC=2*a=6*y. We know that AH=a. AD=x. DH = AD/2 (D center of gravity in AEC). We apply the Pythagorean theorem in the triangle AHC and we have: 6y*6y=3y*3y+HC*HC. Now I apply the Pythagorean theorem in the triangle DHC and we have: 4*4=y*y+HC*HC Subtract the first from the second equation and we have: 36*y*y-16=8*y*y, that is 28*y*y=16. It results that y=2/SQRT(7). But x=2*y so x=4/(SQRT(7)). Q.E.D. Greetings from Romania!

@soli9mana-soli4953 6 ай бұрын

In your first method once drawn point E and knowing that angle EAB is 60 you can apply the external bisector theorem: AC:BC = AD:DB 2a/6=x/2 a=3/2x And then AD^2=AC*AB - CD*DB x^2=3x*3/2x - 2*4 ….or Cosine law on ADB 16=x^2+9x^2 - 2*x*3x*1/2

@prime423 4 ай бұрын

This is an opportunity to talk about the angle bisector theorem. This immediately yields an answer to this problem. Another arrow for a mathlete to use!!

@michaelweiminchang3431 6 ай бұрын

We can use sin formula for triangles ABD and ACD Sin60:BD=SinADB:AB Sin60:CD=sinADC:AC SinADB=SinADC We can get CD=4 Then get middle point E in AC Triangle DCE is similar to Triangle ADC DE:AD=CD:AC We can get AD=x=a, then triangle ADE is equilateral triangle each side is 2 I have thought this problem for some time If it is same as the youtuber , just coincidental

@michaelweiminchang3431 6 ай бұрын

Sorry everyone, I made a mistake. After careful consideration , we can first use cos theorem in triangle in ABC and ADE ,first we get a then we can get x thanks for your time.

@abdmoh6480 6 ай бұрын

Thank you very much for this great math olympiad problem.

@alexutz6450 6 ай бұрын

Hello from Romania!

@MarieAnne. 6 ай бұрын

Here's the method I used: Since AD is angle bisector, then AB/AC = BD/CD → a/(2a) = 2/CD → CD = 4 AD² = AB*AC − BD*CD → x² = (a)(2a) − (2)(4) → *x² = 2a² − 8 (1)* Using law of sines in △ABD 2/sin 60 = x/sin B → *sin B/sin 60 = x/2 (2)* Using law of sines in △ABD 6/sin 120 = 2a/sin B → sin B/sin 120 = *sin B/sin 60 = 2a/6 = a/3 (3)* From (2) and (3) we get x/2 = a/3 → a = 3x/2 Plugging this in (1) we get x² = 2(3x/2)² − 8 2(3x/2)² − x² = 8 9x²/2 − x² = 8 7x²/2 = 8 x² = 16/7 x = 4/√7