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.
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
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 x= 4*sqrt 7/7
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!
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
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!!
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
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.
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
Use sine rule in ∆ABC to get cotB. Calculate sinB. Use sine rule in ∆ABD and get X. No need to calculate "a".
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.
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
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
x= 4*sqrt 7/7
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.
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!
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
the thing is that i’d draw perpendicular lines from D to AC and AB
Thank you very much for this great math olympiad problem.
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!!
Instead of trigonometric ratio we can also use 30-60-90 property also sir
Hello from Romania!
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
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.
what is it derived from 7:50 is it some theorem, is it for random segment or just for a bisector?
Awesome ❤
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
X=4/√7
Nice problem
X^2=16/7 .good days
answer=2x
My computer fell down therefore I can't use it I am writing wiht you on my telephone .
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