One solution with a system of 2 equations with unknown AD AC. From point D draw perpendicular line DE to AC . ED is the half side of AD in the Right triangle ADE(because angle DAE is 30 degrees . Triangle ABC is similar to DEC.( angle ACB is common and angles ABC=CED=90.We have 3 proportional fraction and substituting ED with AD/2 , we have ACAD=4√3 . So we need one more equality with AC and AD . We use the law of the cosines in the triangle ADC for the side DC=2 .We take AC^2 +AD^2=16.From these two equations using the substitution method we take a quadratic equation of unknown AD.We take 2 solutions and we accept that AD=2 .Finally we use the Pythagorian Theorem in the trianngle ABD solving for x side which is 1.
Good solution. Here is one solution with pure geometry: Let E be a point such that ED ⊥ BC and ED = 2√3, and ADE < 90. Then DEC = 30 = DAC, so A, D, C, E are on the same circle, and CE is the diameter of the circle because EDC = 90. So radius of the circle is 2. Let M be the midpoint of CE, then ME = MC = MD = MA = 2. Let H be the food of perpendicular from M on BC. Then DH = 1 and MH = √3. It means MA // BC, which yields BH = MA = 2. Then BD = BH - DH = 2 - 1 = 1.
I worked with the smaller triangle, so that: 1) 1.712² + X² = H² 2) H² - X² = 3 (H + X) * (H - X) = 3 *For this multiplication to give 3, only 3 * 1 = 3, so it turns out: H + X =3 H - X = 1 2H = 4 H = 2 Then the issue is resolved: 4 = 3 + x² X² = 1 X = 1 Bingo!!!!!!!! Bigger triangle will be 3 + 9 = H² 12 = H² H = 3.463 Bingo!
Draw a line DE with E on AC such that ∠ADE = 90°. Drop a perpendicular line EF on BC. ∆ABD is Similar to ∆DFE, hence DF/DE = AB/AD DF = √3*(DE/AD)=√3Tan30=1 Hence DF = FC = 1 and ∆EFD is Congruent to ∆EFC. ∠DEC = 120° and therefore ∠DEF = ∠FEC = 60° = ∠BAC Hence ∠BAD = 30° and X = 1
A more general solution for this kind of problem is: After construct triangle DEF, we can know the length of DF and the value of BD:EF, then use EF:AB=CF:CB to get BD. In this way, as long as angle DAC is special angle(Angle BAD and ACB can be any angle), we can solve it with pure geometry method.
One solution with a system of 2 equations with unknown AD AC.
From point D draw perpendicular line DE to AC . ED is the half side of AD in the Right triangle ADE(because angle DAE is 30 degrees . Triangle ABC is similar to DEC.( angle ACB is common and angles ABC=CED=90.We have 3 proportional fraction and substituting ED with AD/2 , we have ACAD=4√3 . So we need one more equality with AC and AD . We use the law of the cosines in the triangle ADC for the side DC=2 .We take AC^2 +AD^2=16.From these two equations using the substitution method we take a quadratic equation of unknown AD.We take 2 solutions and we accept that AD=2 .Finally we use the Pythagorian Theorem in the trianngle ABD solving for x side which is 1.
Good solution.
Here is one solution with pure geometry: Let E be a point such that ED ⊥ BC and ED = 2√3, and ADE < 90.
Then DEC = 30 = DAC, so A, D, C, E are on the same circle, and CE is the diameter of the circle because EDC = 90. So radius of the circle is 2.
Let M be the midpoint of CE, then ME = MC = MD = MA = 2.
Let H be the food of perpendicular from M on BC. Then DH = 1 and MH = √3. It means MA // BC, which yields BH = MA = 2. Then BD = BH - DH = 2 - 1 = 1.
I worked with the smaller triangle, so that:
1) 1.712² + X² = H²
2) H² - X² = 3
(H + X) * (H - X) = 3
*For this multiplication to give 3, only 3 * 1 = 3, so it turns out:
H + X =3
H - X = 1
2H = 4
H = 2
Then the issue is resolved:
4 = 3 + x²
X² = 1
X = 1
Bingo!!!!!!!!
Bigger triangle will be
3 + 9 = H²
12 = H²
H = 3.463
Bingo!
Got it!
Thanks.
Nice solution. Hard international math olympiad problems in the future?
Did you solve this?
Si E pertenece a AC y DE es perpendicular a AC》Ángulos: D=180°=60+60+60 ; DCE=30° ; BAD=30°》Si DC=2》AD=2》BD=1
Gracias y saludos.
Tres bien
φ = 30°; ∆ ABC → AB = √3; BC = BD + CD = x + 2; CAD = φ; x = ?
AC = AE + CE → sin(DEA) = 1 → BD = DE = x → EAD = DAB = φ → x = √3/√3 = 1
You could have canceled the 3x with the 3x.

Draw a line DE with E on AC such that ∠ADE = 90°. Drop a perpendicular line EF on BC.
∆ABD is Similar to ∆DFE,
hence DF/DE = AB/AD
DF = √3*(DE/AD)=√3Tan30=1
Hence DF = FC = 1 and ∆EFD is Congruent to ∆EFC.
∠DEC = 120° and therefore
∠DEF = ∠FEC = 60° = ∠BAC
Hence ∠BAD = 30° and X = 1
Why is ∆ABD similar to ∆DFE ?
@@howardaltman7212 Since ∠ADE = 90°, ∠ADB + ∠EDF = 90°.
Hence, ∠BAD= 90-∠ADB =∠EDF, and similarly, ∠ADB= ∠DEF
A more general solution for this kind of problem is: After construct triangle DEF, we can know the length of DF and the value of BD:EF, then use EF:AB=CF:CB to get BD. In this way, as long as angle DAC is special angle(Angle BAD and ACB can be any angle), we can solve it with pure geometry method.
@@幕天席地-w9c:
Great method!
no， I don‘t see it. it's easy to prove you are mistaken. draw it on any right-triangle, not this one.
It turned out that both triangles ABC and ABD are not only similar but also 30°-60°-90°.
Easy
X=1
Ser plss
Gud evning ser plss gibme nbr 3d. 4d