Hard Geometry Problem - Contest In Switzerland

[@MindYourDecisions]

I always get many emails for problems like this! Sorry I don't have time to review alternate methods/proofs, so if you don't see a reply from me that is why! I do encourage people to prepare their own videos, and lately I've been sharing such works on Twitter @preshtalwalkar and on KZbin. Great to see the excitement for a problem like this one!

[@teresamiles5929]

120

[@aashutoshsharma8476]

how to send u question

[@lifeofphyraprun7601]

Smtimes I also hav problems which I wish to send to u.How can I send those?

[@lifeofphyraprun7601]

Well,Happy Belated Teachers Day(it was 5 days ago,but still)my best math teacher!I thought that telling u in a personal reply wd be better than in a public comment.Anyways,I also wished u on 5th September(Teachers Day)on another video(probably ur previous one).

[@abdixsimplix2582]

Can you write your e-mail here please?

[@spiderjerusalem4009]

"Hard Geometry Problem" But answered trigomonetrically.......

[@franciscotrigo9656]

But with no calculator! I don't think this problem could ever be solved without trigonometry, Could it?

[@TechToppers]

@@franciscotrigo9656 It can be... The point is, it will require a lot of geometric constructions... And lot of equation solving.

[@rtfacts5317]

@@franciscotrigo9656 i can solve it without trigonometry.

[@anurag11-b66]

@@rtfacts5317 👍shud do it....

[@gregheffley6400]

@@rtfacts5317 post a video then

[@Jop_pop]

There is a much simpler solution: Extend AC to a line. Draw a perpendicular to the line so that it intersects point B. Call E the intersection of the perpendicular with the extension of AC. The triangle ABE is a 30-60-90 triangle by supplementary angles. So AE is half of AB. Thus AE=AC. Adding this to AC we get CE=2AC. Moreover, BE=sqrt(3)*AE by 30-60-90. Note BCE is a right triangle. Use the pythagorean theorem to get BC^2=CE^2 + BE^2 =4(AC)^2 + 3(AC)^2 =7(AC)^2 Therefore BC=(AC)sqrt(7). EDIT: See my comment below where I finish the argument to get it in terms of AD

[@Chriib]

Now, that is a very smooth and nice solution. No trig functions needed, lovely.

[@aarontcao]

You still need trig to solve for AC at the end. The problem is asking for a length so you can't figure out 150sqrt(7) without the law of cosines.

[@Asdfgfdmn]

Nice way but you still have to figure out AC

[@Jop_pop]

@@aarontcao You're right I didn't do that part (forgot we were given AD not AC) but here is a way to circumvent that with elementary geometry: Draw a line through D parallel to AB. Call its intersection with AC point F. By alternate interior angles you have a 60° angle ADF. So ADF is an equilateral triangle. Then by supplementary angles, angle CFD is 120°. But that's the same as CAB. Thus triangle ABC is similar to triangle FDC since they share that angle and also angle ACD. Do the same thing on the other side: draw a line through D parallel to AC. Call its intersection with AB point G. By the same argument, ABC is similar to GDB. Thus GDB is similar to FDC. But length GD=FD by equilateral triangles. And segment GD corresponds to segment FC in similarity between triangles GBD and FDC. And FD=2(FC). So the ratio of similarity between FDC and GBD is 2. Therfore BD=2(DC). So BC=3(DC). So the similarity ratio of ABC to FDC is 3. Therefore AC=(3)FC. But FC=AD/2 so AC=(3/2)AD. Combine with the previous result and we're done.

[@leif1075]

WAIT A MINUTE ISNT THE ANGLE ON EITHER SIDE,OF POINT D 90 DEGREES SO,ANGLE ADC AND ANGLE ADB ARE 90 DEGREES?

[@oldguydoesstuff120]

I've forgotten way too much trig to solve this one.

[@mffbataineh]

This is my solution to the same problem - I believe it is nicer than the one presented. kzbin.info/www/bejne/omWWYmCBqN1jqpI

[@shubhendranathsingh9888]

You could obviously use sine rule in both triangles and could say BD,CD relation with angle bisector theorem. Taking both angles angle B,C as 30-x and 30+x respectively. Easily 2 equations. One with trigonometry and then substitute in other to get the answer.

[@shubhendranathsingh9888]

@ME - 12ZZ - AEC South (2392) yeah like everything will come from Stewart theorem. You only know one side remember that. Stewart theorem is just extension of cos rule and noway it works with only 1 known side.

[@shubhendranathsingh9888]

@ME - 12ZZ - AEC South (2392) I should say can you explain me how?

[@zeeve284]

No trigonometry is required here at all! Just pythagorean theorem and basic geometry!

[@PranavKumar-ns8cr]

Hi fresh lakewater

[@MohammedJabedAhmed-ec4zs]

😂

[@iMvJ27]

😂 😂

[@denvercheddie]

Alternative method. Define a point E on AC such that ADE is equilateral with all sides = 100. You will see that triangles BAC and DEC are similar, and therefore AE = 100, EC = 50 and x = 100+50.

[@Railgo7]

nice one, thank you

[@pinkikumari-vb4px]

I have done like this

[@TengouX]

Draw a line to extend BA from point A onwards, then draw a line up from C parallel to AD. You now have an equilateral triangle spanned by point A, point C, and the new point E. Using equal angles and one single application of the law of cosines, you can now calculate BC

[@ExpressStaveNotation]

Nice

[@MrRyanroberson1]

First we calculate some areas. The areas of the triangles ABD, ADC respectively: AB * AD * sin(60)/2, AD * AC * sin(60)/2, by composition: ABC measures (AB+AC) * AD * sin(60)/2, which evaluates to 150 sqrt(3/4) AC by the previous formula: ABC measures AB * AC * sin(120)/2, which evaluates to sqrt(3/4) AC^2 as these areas are equal, we can divide everything by sqrt(3/4) AC to get 150 = AC finally we use the formula for the side BC: BC^2/AC^2 = 1 + 4 - 4cos(120); cos(120) = -1/2 and so we get that BC^2/AC^2 = 7. BC = 150sqrt(7)

[@itsahmd295]

I thought about this directly when I opened the video

[@sleepyowl6982]

Nice! I used a different method to solve this question. From the angle bisector in a triangle theorem, AB:AC = BD:CD=2. Let CD=y, BD=2y and AC=x AB=2x. From D draw a line parallel to AC and intersect AB at E. Angle BED=120, so angle AED=60. From given angle BAD=60 and angle AED=60, triangle AED is an equiangular (equilateral too) triangle so DE=100, AE=100. Triangle BED is similar to triangle BAC, BE:BA=BD:BC=2:3, (2x-100):(2x)=2:3, x= 150=AC, so AB=300, BE=200. Use law of cosine in triangle EBD(the numbers are easier to manage), BD=100 square root 7, then CD= 50 square root 7. BD+DC=BC=150 square root 7.

[@HotelPapa100]

You can even solve this without any trigonometry, except from values known from special triangles With angle BAC 120° you can take this triangle as cut off from an equilateral hexagon with one side of the hexagon AB, and AC half of the next.You can then extend AD to the center of the hexagon, lets call this center E. AB=BE=AE. Triangles ADC and BED are equiagular, so AD/DE=AC/BE. Some wrangling of this equation and plugging in AD= 100 gives us AE=radius and side of the hexagon = 300. CE is the height of the equilateral triangle ABE, so 1/2AB x sqrt(3). Pythagoras gives us BC = sqrt (300^2 + 150^2 x 3) = sqrt (4 x 150^2 + 150^2 x 3) = 150 x sqrt(7)

[@mophiesez]

Nice solution. And I have another way to solve it: extend BA to point E, making AE=AC, then connect EC. Since ∠CAE=180-120=60 degree, and AE=AC, △ACE is an equilateral triangle. Therefore, ∠AEC=60 degree. Then it turns out that AD∥CE, so CE/AE=BE/BA=3/2, AC=CE=150, AB=2AC=300. With law of consines, we have BC=150sqrt(7)~~

[@happyhappyguy5034]

i think u mean CE/AD instead of CE/AE

[@Willd2p2]

I think there is a typo here: CE/AE=BE/BA=3/2. CE and AE are two lines in the equilateral triangle ACE so the ratio can't be 3/2. I think it was supposed to be CE/AD=BE/BA=3/2.

[@happyhappyguy5034]

@@Willd2p2 wow we comment this in the same time :)

[@warfyaa6143]

Identical to my sol.

[@shiva_gaming4997]

Please solve, a+b+c=0, a²+b²+c²=42 , a³+b³+c³= 105, then show that,, (a-b)(b-c)(c-a)=±63

[@vusalburcaliyev7153]

I am from Azerbaijan and when I saw Tusi on your video, I was too glad Because in our school we are just teached the law of Sines, we are not told anything about Tusi even though he is from Azerbaijan

[@vusalburcaliyev7153]

@@MehrMoon1335 Yes, you are right, and If I am not mistaken these areas were under the control of Atabaylar(Eldanizs) which was a Turkish(nationality) country

[@krishnavpai]

We can set up the coordinate axis, X Axis along AB. B(2,0) and get C as (-1/2,√3/2). Then finding the coordinates of D by solving the lines y=√3x and y=-√3/5(x-2) (pt. slope form) D(1/3,1/√3) then dist formula, we get AD as 2/3. Now scale up(zoom) the triangle such that AD is 100 (2/3*150). BC is found to be √7. Thus answer is 150√7. Hope you like it

[@notsoancientpelican]

I also used coordinate geometry. It's often the quickest and surest solution.

[@yurenchu]

@Krishna Pai, if you set up the coordinate axes such that x-axis is along AD, and let |AC| = a, then you don't even have to write line equations.

[@shelleyweiss9920]

In addressing this problem I asked myself where I might recognize this figure. I viewed it as a section of a regular hexagon (with AB as one side, and point C as the midpoint of the adjacent side), or simply as part of a 60/120/60/120 rhombus (again, with point C as a midpoint on a side). By constructing and subdividing several right triangles into the figure, I eventually stumbled my way to showing that BC=(SQRT 7)*AC, and that AC = (3/2)*AD. Upon reading the comments, however, I was much more impressed with Joe Previdi's more elegant solution using equilateral triangles (thank you Joe!)

[@atharv_008]

I am Atharv and I study in 10th grade in India, and I don't find this problem much difficult as I sometimes solve Olympiad level kind of problems. I solved it completely using pure EUCLIDEAN GEOMETRY. Let's look at the problem now! Since, AB=2AC and AD is the angle bisector ----> By angle bisector theorem, we get BD=2DC. Draw a line parallel to DA through B. Let the parallel line intersect extended CA at E. Now, using basic properties of parallel lines and transversals, we get traingle ABE an equilateral triangle. Triangles ADC and EBC are similar by AA corollary. Also, since AD=100 and AB=2AC, we get AC=150 and AB=AE=EB=300. Draw DM perpendicular to AB such that M is on AB. By 30-60-90 triangle theorem, we find AM=50 and DM=50√3. By Pythagoras theorem in triangle BMD, we find BD=100√7 ----> DC=50√7. Hence, BC=150√7.

[@vaishalitirthkar]

This is a simple and easy solution.

[@atharv_008]

@@vaishalitirthkar Thanks.

[@AugustinJacobvadakayil]

Just use triangle congruence: 2AC =AB, angle BAD = angle DAC therefore they are congruent triangles. Implies AC/AB = CD/BD, IMPLIES BD = 2DC IT'S THAT EASY.

[@matthewbusche4547]

Here's a way to do it where the only trig you need is to know the sin/cos of a 30 degree angle. To eliminate the need for variables, start by changing the problem so |AB| = 2, |AC| = 1 and |AD| is instead unknown. (With the intention of scaling things up once you find |AD|). Then rotate the picture so A is at the origin, and point B is at (0, -2). Noting that point C is then at (sqrt(3)/2, 1/2), you can immediately write down: |BC| = sqrt((5/2)^2 + (sqrt(3)/2)^2) = sqrt (28/4) = sqrt(7) It's also easy to write the equations for line (A, D) and line (B, C) in slope-intercept form: y = -x / sqrt(3) y = 5 / sqrt(3) - 2 The intersection of these two lines is at point D. Solving you get D = (sqrt(3)/3, -1/3), so |AD| + sqrt(3/9 + 1/9) = sqrt(4/9) = 2/3 So we need a scaling factor or 150 to get |AD| up to the given value of 100, which means (for the actual given problem) |BC| = 150 sqrt(7). Perhaps not elegant, but less work than what Presh did, and you don't need to know much this way either.

[@navaratnammanoharan8332]

From D draw angle 60 deg. from DA to meet AB at E. ADE is equilateral. DE=EA=100. AE = 2x-100. 2x-100/2x = 100/x. x=150. BC^2= 300^2+150^2-2*300*150*cos120=150*square root 7

[@riteshbhartiya6155]

Easy peasy! { Using only- values of sin60°,cos 60° , slope of line, distance formula } Take A(0,0) as origin of a co-ordinate system. Take B in first quadrant and C in fourth. Also, D(100,0) lies on X-axis. Since the angles are given, find points, B(AC,√3 AC) and C(1/2 AC, - √3/2 AC). We have found points B and C, now find slope of BC=3√3 Since D(100,0) lies on BC, (√3AC-0)/(AC-100) = 3√3 => AC = 150 using distance formula, BC = √[ (1/2 AC)² + (3√3/2 AC)² ] => BC = AC √7 => BC = 150√7 ✓

[@mogewszystko3896]

I suggest a simplier way. AC=x, AB=2x. The area is equal to 1/2*2x*x*sin120°. It can be counted based on two little triangles, 1/2*100*2x*sin60°+1/2*100*x*sin60°. Then we have x=150. Then BC^2=(2x)^2+x^2-2*x*2x*cos120°. Then we have an answer

[@armlessjohn666]

I tried a couple of problems from this channel before and had no luck. This is the first that I got correct!

[@amedlolo7375]

150*(root)7, put AB=2k,AC=k, area of triangle ABC =2k*k*sin120*.5=k*100*sin60*.5+2k*100*sin60*.5 OK, so k=150, put BC=j, by using cosine law, cos120=4k^2+k^2-j^2/2*2k*k so 2.5-j^2/2k^2=-1 so j^2=k^2*7 so j=k*(root7) =150 (root 7)

[@jovangerbscheid4619]

AB is an angle bisector of angle DAC and AC is an angle bisector of angle DAB, so you can use the angle bisector theorem to find that AD/AC=BD/BC=2y/3y=2/3, so AC=3AD/2=150

[@illiil9052]

extend AC towards point C by its length to get point E. let line AD meet line BE on point F. it is quite easy to find out point D is center of gravity of triangle ABE, so AF=1.5*AE=150. triangle ABF is 30-60-90 triangle, so length of AB=2*AF=300. use cosine rule for BC^2=AB^2+AC^2-2*AB*AC*cos(BAC) to get BC.

[@shiva_gaming4997]

Please solve a+b+c=0, a²+b²+c²=42 , a³+b³+c³= 105, then show that,, (a-b)(b-c)(c-a)=±63

[@trouthuang1959]

Extend line BA to point E so that line EC is parallel to line AD. Angle BAD = Angle BEC = 60 degrees. Also that Angle EAC is 180-60-60 = 60 degrees. So that triangle AEC is an equilateral triangle. Set line AC = a = line AE = line EC, and line BA = 2a. By similarity, 2a : 100 = 3a : a, which shows a = 150. Line AC = 150, and line AB = 300. Then you can get line BC by using trigonometry in triangle ABC.

[@anandk9220]

It may sound unbelievable. But I actually managed to solve this one orally in about 15-20 minutes. Just used basic geometry of isosceles triangle, property of angle bisector sides and cosine rule. Here's the solution - Simply draw CM to midpoint M of AB. Then use basics of isosceles triangle and keep applying cosine rule to find every other side in terms of AC. Also use angle bisector property to find BC and DC in terms of AC. Finally apply cosine rule in triangle ADC to find quadratic equation giving two values of AC (out of which 300 is impossible as altitude of triangle AMC through A will be shorter than AD). Hence AC = 150 So, BC = √7 × AC = 150√7 units

[@vladimirrainish841]

150*sqrt(7) . Extend AC beyond C on the length of AC, the resulting triangle BAM will be isosceles triangle where AB = AM = 2AC , BAM angle is 120 degrees, ABM = AMB = 30 degrees. BC will be a median in BAM triangle , since AC = CM. AD will also be a median (and a height ) and since D is the point where two medians intersects and AD = 100, then DN = 50 ( N is point where extended AD meet BM side of new triangle) it's half of AD which is 100, intersecting medians theorem says medians divides each other 1:2 ratio) , so AN = 150. if AN = 150 , then AB = AM = 300 (150/sin(30)) ( since ABM = AMB = 30 degree) and BN = NM = 150*sqrt(3).By Pythagore theoreme BD = sqrt(50^2 + 3*150^2) = sqrt(50^2 +27*50^2) = 50*sqrt(28) = 100*sqrt(7) , since it's 2/3 of BC , BC = 150*sqrt(7). Not sure about computations(put it in hurry) , but median approach makes it simple. Let me also, add that this solution doesn't use any trygonometry;: median theorem is proven using only similarity/identity of triangles, and the fact that smaller leg of 30/60/90 triangle is half of hypotenuse can be proven using only triangle identity as well

[@richfi9576]

I solved it using the 30-60-90 triangle theorem, similar triangles & Pythagoras, no sine/cosine rules required. It took me a lot longer than the length of this video though.

[@Railgo7]

honestly, i like this much more than the one used in the video :D

[@sergten]

Another approach is to use the formula for the area of the triangle as S = 0.5 * sideA * sideB. So S0 = 2x*x*sin(120), S1 = 2x*100*sin(60), S2 = x*100*sin(60), S0 = S1 + S2 => x = 150. Then the cosine theorem gives y^2 = (2x)^2 + x^2 - 2x*x*cos(120) = x * sqrt(7).

[@akashsudhanshu5420]

Use bc[1-(a/b+c)²]=AD² for angle bisectors And then cosine law for 120°

[@d.m.7096]

Solution using pure geometry - By Angle Bisector theorem, AB/AC = BD/DC = 2/1 Let AC = x. Then AB = 2x. Hence, let BD = 2y and DC = y. Now, AD^2 = AB.AC - BD.DC Hence, 10000 = 2x^2 - 2y^2 Hence, x^2 - y^2 = 5000 Draw DE perpendicular to AC. Hence, ∆ADE is 30-60-90 triangle. Therefore, AE = 50 and DE = 50√3 and EC = x - 50. By Pythagoras theorem for ∆DEC, y^2 = (50√3)^2 + (x - 50)^2 Hence, y^2 = 7500 + x^2 + 2500 - 100x Hence, x^2 - y^2 = 100x - 10000 = 5000 This gives, x = 150 and y = 50√7 Hence, BC = 150√7

[@christianfunintuscany1147]

With Carnot theorem first calculate BC in function of x, you find BC= x sqr(7) Then you draw two lines from D: one parallel to AC and the another parallel to AB. You get three simile triangles from which: BD = (100/x) BC = 100 sqr(7) CD = (50/x) BC = 50 sqr(7) So you get BC = 150 sqr(7)

[@rasmusturkka480]

I did it by solving BC in terms of x by cosine law to get BC = sqrt(7) * x and then applied the equation Area of ABD + Area of ACD = Area of ABC by using the sine formula Area = 1/2 * ab sin(theta) to solve for x = 150.

[@petermitsis685]

If you make a mirror of triangle ADC along line AC, you get one large triangle (BCD') made up of three smaller triangles.The areas of each of the small triangles can be found in terms of AC, and their sum is the large triangle BCD'. The area of BCD' is also equal to (1/2)*(BD')*(AC)*sin(60). Solve for AC. Now use the cosine theorem to solve for BC.

[@nishantchhonker2772]

I actually used concepts of similar triangles. I draw a line parallel to DA that will intersects with AB produced ( say at E) . Then ∆ABD ~∆ ACE. Also ∆ACE is an equilateral triangle.

[@ExpressStaveNotation]

Nice. Except you didnt say to draw the line from C! Then its the same as Tengou's comment above.

[@Kali-tl5uc]

Other solution. By the bisectriz theorem BD=2DC; let F be the mid point of AD and draw the line parallel to AB passing through F, this line intersects AC at G. It is straightforward to show that triangle AFG is equilateral. Also, by using the cosine of 30 degrees one gets that AD=2AG. From this, one obtains that AC=3AG=(3/2)AD. On the other hand, by using the cosine theorem, one gets BC=sqrt(7)AC, thus BC=(3sqrt(7)/2)AD.

[@a_man80]

Before wacthing this video, I have already found a formula for angle bisector of 120°. formula: AD=AB×AC/(AB+AC) AD=100 , AB=2AC=2x , AC=x 100=2x²/3x=2x/3 , x=150 , 2x=300 with cosine theorem we find BC=150√7 To prove this formula I wrote 3 cos theorem ( two for 60° angles and one for 120°) and angle bisector length theorem (AD²=AB×AC-BD×DC) combining these four equations (left an exercise for reader) we get this formula.

[@pbierre]

I used Cartesian coords, line equations and line intersection points. First, mirror and rotate the outer triangle so that: A is at the origin D is at [ 100, 0 ] B is at [ s , s*sqrt3 ] (s is unknown length of AB) C is at [s/2, -s*sqrt3/2 ] (s/2 is unknown length of AC) L1: The line equation of line BC is: y = sqrt27 * x - 100*sqrt27 (using 2-point formula for slope, s cancels out) (use known point [ 100, 0 ] to get y-intercept) L2: The line equation of line AB is: y = sqrt3 * x L3: The line equation of line AC is: y = -sqrt3 * x B is the intersection L1&L2. C is the intersection of L1&L3. Once these are solved, the answer is the distance BC between these 2 intersection points = sqrt(157,500) = 396.862

[@bobzarnke1706]

The cosine law for ΔABC gives: BC² = (2x)² + x² - 2x²cos(120°), which reduces to BC = x√7. Since ΔABC = ΔABD + ΔADC, computing the areas gives: 200x sin(60°)/2 + 100x sin(60°)/2 = 2x²sin(120°)/2, which reduces to, x = 150.

[@mmattoso1]

I did as following: We see (area of ABD) = 2.(area of ADC) (from the area formula using (sin60°) and both adjacent sides to the angle) Therefore, (area of ABC) = 3.(area of ADC) If AC=y, AB=2y, and we will have: (1/2).2y.y.sin(120) = 3.(1/2).y.100.sin(60) From where we get y=150 At this point, we can use cossines law for the triangle ABC, if BC=x we get: x^2 = (2y)^2 + y^2 - 2.(2y).y.cos(120) From where we get x = y. sqrt(7) And as we already know y, we'll get the answer: x = 150.sqrt(7)

[@artecuartico]

Extend BA and create an equilateral triangle AEC, triangles ABD and EBD are similar, so BA/AD = BE/EC, then EC = CA = AE = 150, use cosines law (or geometry) to find DC = 50 sqrt(7) and BC = 3DC = 150sqrt(7).

[@sswy1984]

you can do BM CN both perpendicular to AD. BM=2CN. AN=1/2 x. DM=2DN. so 1/2x *(1+1/3)=100. then x = 150.

[@RanjanParajuli]

another way to do this is to equate (triangleABD+triangleADC) = triangleABC using sine formula for area (remember area of triangle = 1/2sinA bc), you will directly get AC=150 and after that just use cosine law for triangle ABC and you will get BC

[@demasmoha]

Can you do it practically

[@avikdas4055]

Presh Talwalker: So let's use Law of Sines, Law of Cosine, Trigonometry............ Stewart: Am I a joke to you?

[@avikdas4055]

@@williamzhang8659 lol

[@equbalfatmi860]

Yeah but after using Stewart's theorem we have to use law of cosines to get the answer

[@avikdas4055]

@@equbalfatmi860 Yes....... After using ABT u use Stewart and u will be almost done. The rest is instinctive with basic trigonometry and algebra.

[@saxena.apoorv]

Very nice!

[@SimpleTeam1975]

This problem is easy. If you draw a parrel line from D, which is parrel to AC, and cross AB at point E. The triangle ADE is a special triangle. and triangle BDE is similiar to ABC. The length of AB is easy to be calculated.(100 + 200) and AC is 150.

[@nandakumarcheiro]

Thank you for reminding Al Kashi theorem wonderful.When applied in a right angled triangle -2x l cos theta becomes zero at cos 69 becomes cos 90 being zero this term vanishes to zero reinforcing Phythogaress hypotnuse square equivalent to sum of side squares.Reminding us for sine law after a long peiod.

[@neshploda17]

To find AC, I graphed it on some spare trisymmetric grid paper I had lying around. I noticed that AC was 1.5x the length of AD, or 150. And then from there it was just Law of Cosines to find BC. I knew that trisymmetric grid paper would come in handy!

[@tamonekicofi]

Extend AC to a line. Draw a line from B that intersects AC at point E, such that angle EBA equals 60. since angle BAE equals 60, triangle BAE is equilateral with a side of AB=2AC. Also Triangles BCE and DCA are similar, so 100/AC=BE/EC=2/3. Then we get AC=150 and by the law of cosines BC=150sqrt(7)

[@giacomolanza1726]

Another simple solution without trigonometry. We can complete an equilateral triangle ABP by prolongating the bisector AD. BD = 2 CD by the bisector theorem and consequently PD = 2 AD (due to Thales's theorem, or to the similarity of triangles DCA and DBP). Then AP = 3 AD = 300. The constructed equilateral triangle has all sides equal to 300 and height CP = AP sqrt3 / 2 = 150 sqrt3. By Pythagoras' theorem then BC = AP sqrt7 / 2 = 150 sqrt7.

[@HuongTran-rm4fd]

From C draws a paralel line that intersects AD at E.

[@shiva_gaming4997]

Please solve a+b+c=0, a²+b²+c²=42 , a³+b³+c³= 105, then show that,, (a-b)(b-c)(c-a)=±63

[@henrilaporte7599]

Put the origin at A and rotate the axis 180 degrees. Some notations: s = sin(60) , c = cos(60) , t = tan(60). D = 100 * ( u , v) Where u**2+v**2 =1 and v < 0. C L * (c*u - s*v , s*u + c*v) Rotation 60 degrees of (u,v), L unknown B = 2*L * (c*u +s*v , -s*u + c*v) Rotation -60 degrees of (u,v), L unknown Because B , C and D are on the same y coordinate, we find L*c = 75 and v = 3*t*u. From the last equation and from u**2+v**2 = 1 and t**2 = 3 , we got u**2 = 1/28 and u = - 1 / SQRT(28). Let replace v by 3*t*u , factorize c*u and after replace L*c by 75, u by 1/SQRT(28) and t**2 by 3 C = L * c * u * (1- 3*t**2 , 4t) = 75 * ( 8 , -4t) / SQRT(28) B:= 2 * L * c * u * (1+3*t**2 , 2t) = 75 * (-20 , -4t) / SQRT(28) Distance between B and C = 75*SQRT(28) = 150*SQRT(7)

[@smchoi9948]

"Tan" could convert angles to slopes and turn the problem into a co-geom. one: Let |AC| = s. Align the figure on the Cartesian plane so that C & A are at C(0,0) & A(s,0). There is a line L, w/ eqn. y = mx for some m, that passes thr'u C, D & B, so the last 2 are at D(q,mq) & B(p,mp) for some q>0 & p>0. The eqn. of the line containing AD is y = tan (180°-60°) x + u where 0 = -√(3) s + u, i.e. √(3) x + y = √(3) s; as D is on it, √(3) q + mq = √(3) s s = q + mq/√(3) ...(i). The eqn. of the line containing AB is y = tan [180°-2(60°)] x + v where 0 = √(3) s + v, i.e. √(3) x - y = √(3) s; as B is on it, √(3) p - mp = √(3) s s = p - mp/√(3) ...(ii). As |AD| = 100, (s-q)² +(0- mq)² = 100² [mq/√(3)]² +m²q² = 100² (by (i)) m²q² = 7500 ...(iii) As |AB| = 2|AC| = 2s, (s-p)² +(0- mp)² = (2s)² [-mp/√(3)]² +m²p² = 4s² (by (ii)) m²p² = 3s² ...(iv) m²p² = 3[p - mp/√(3)]² (by (ii)) m² = [√(3) - m]² √(3) - m = ±m (-ve branch is rej.) m = √(3)/2 ...($) (iv) / (iii): p²/q² = s²/2500 p/q = s/50 ...(*) (N.B. p/q>0) Equating (i) & (ii), q + mq/√(3) = p - mp/√(3) [√(3)+m] q = [√(3)-m] p [3√(3)/2] q = [√(3)/2] p (by ($)) p/q = 3 s/50 = 3 (by (*)) s = 150 ...(¥) By (iv), (¥) & ($), p = √(3)(s/m) = 300 ...(£), so |BC| = √[(p-0)² + (mp-0)²] = p√(m²+1) =(300)[√(7)/2] (by (£) & ($)) =150√(7).

[@EdjarbasOliveiraJunior]

Let AC=x and CD=y.Then 2x•x/2x+x =100 , x=150. Lenght bissector: 100•100=2x•x-2y•y , y=50√7.Then BC=3y=150√7.

[@giffmimi]

Extend AD, AC. Drop perpendiculars C to AD, B to AD, B to AC. Use ~ triangles, 30, 60 Rt Triangle stuff and Pythagorean theorem.

[@jimlocke9320]

Let length AC = x, then AB = 2x, and let length CD = y, then BD = 2y by the angle bisector theorem and AB being twice as long as AC. Extend AB at A and drop a perpendicular to it from C, labelling the intersection as point E.

[@andreasask6791]

I completed the picture with 60-30-90 triangles to the left and right of A, using the existing 60 angles. The uniformities could then be used together with Pythagora theorem to solve this.

[@nianli4428]

From D to draw a parallel line to AC, cut AB at E, from two similar triangles, it easy to get AC equals 150 then AB is 300.

[@zeeve284]

Exactly!!!

[@robertocaesar]

This is easily solved without Trigonomertry by drawing one equilateral triangel with the angle

[@pi17]

Hey why don't you just use the Stewart's theorem? That will solve it in in secs

[@joshspektor4389]

Piyush Mohite won’t solve it right away since the lower side has no parts given. you’d still need to use law of cosines on one of the triangles to solve the system

[@miguelangelyc4439]

Cuaterna armonica nomas

[@charnson247]

YES STEWART'S THEOREM ROCKS. LoL...MAA thinks we don't know this bashy theorem haha.

[@ayynow5128]

It wont work, the base isnt divided into three parts ando so stewarts theorem would complicate it more

[@nishatiwari9212]

I tried it using Stewart but there isnt enough information in question to solve it fully.

[@figurine8122]

(1) Extend BA to BE such that AE = x. (2) ∆AEC is equilateral (AE = AC = x and

[@kthunsicker]

I really enjoy your solutions to these challenging problems. I am a retired math teacher.

[@puar07]

thanks for teaching the youth of this world, happy retirement

[@renesperb]

An interesting problem is to construct this triangle with straight edge and compass . It also requires different concepts to be applied.

[@warfyaa6143]

alternative solution : Extend AB from A to E such that AE = AC angle CAE = 60 so triangle ACE is equilateral moreover triangles ABD & BCE are similar from similarity, BE=3AC=(3/2)BA, so AC=(3/2)AD = 150 using Law of cosines for triangle ADC ==> DC=50 * sqrt(7) from similarity, BC=3DC = 150 * sqrt(7)

[@shiva_gaming4997]

Please solve a+b+c=0, a²+b²+c²=42 , a³+b³+c³= 105, then show that,, (a-b)(b-c)(c-a)=±63

[@warfyaa6143]

@@shiva_gaming4997 hmm, this is little bet advanced algebra .

[@shiva_gaming4997]

@@warfyaa6143 yaa

[@shiva_gaming4997]

@@warfyaa6143 but, how can I solve this problem.

[@giuseppebassi7406]

I did the same, but until 2:43 you could use the bisector theorem because AD is a bisector of BAC. This theorem tells us that AB:AC=BD:DC

[@shadrana1]

By the angle bisector theorem,BD=2DC Let BD=2y,DC=y, and AB=2x,AC=x. Extend BA and meet a perpendicular line from C at E Triangle AEC is a 90,60,30 triangle,AE=x/2,EC=(sqrt3/2)x and AC=x We now do a Pythagoras on triangle BEC, (BA+AE)^2+(EC)^2=BC^2 (2x+x/2)^2+(sqrt(3)/2*x)^2=(2y+y)^2 25/4(x)^2+3/4(x^2)=9y^2 (28/4)(x)^2=9y^2 7x^2=9y^2...……………………….(1) Draw line FD perpendicular to AB, Triangle FAD is a 90,60,30 triangle,DF=(sqrt(3)/2)100,FA=100/2and AD=100. We now do a Pythagoras on triangle BFD, BF^2+FD^2+BD^2 (2x-50)^2+(sqrt3/2*100)^2=(2y)^2 4x^2-200x+2500+3/4(100)^2=4y^2 4x^2-200x+10,000=(28/9)x^2 (substitute from (1)) (8/9)*x^2-200x+10,000=0 8x^2-1800x+90,000=0 x^2-225x+11,250=0...………..(2) (x-150)(x-75)=0 x=150 or 75. Try x=150, 7*(150)^2=9y^2...…………………….from(1) Take sqrt of each side. y=150/3*sqrt7=50sqrt7. BC=3y=3*150/3*sqrt7=150sqrt7 as required AD=100>75,therefore x=75 does not exist in the real world.(AB and AC too short to fit.)Maybe there is another shape that this algebra fits into but I can't find it.

[@JohnJones-pu4gi]

As a matter of little interest, this figure is known as the optician's nomogram. This is because it solved 1/u + 1/v = 1/f in the days there weren't any apps. Here 1/AB + 1/AC = 1/AD. Given the constraints we get immediately AC=150, AB=300 and then stuff it into the cosine rule and bob is your aunt's spouse. HTH

[@aqibzaman7831]

Saw the thumbnail: oh this will be easy. After watching video: Wait what? NO!

[@davidgracely7122]

Draw a line from point D to a point we'll call E on line segment AC so as to make triangle ADE an equilateral triangle. Triangle DEC is a similar triangle to triangle BAC and thus has the same trigonometric ratios as triangle BAC. Thus line DE is 100 and EC is 50 which when added to AE makes side AC = 150. This means that side BA = 300. Then use the Law of Cosines to compute the length of BC. In geometry, the problem is to know how much Euclidean geometry to use and how much analytical geometry to use in order to come to the answer with the least amount of moving parts. One of the strange things about math is that sometimes there is more than one way to solve a problem. That to me is refreshing, because none of us have brains that are wired exactly the same. I really think that sometimes the Lord may anonymously bring a thought to a person's mind to solve a problem because I was lost in a morass of too many variables and not enough equations when suddenly drawing just this one little line unlocked the problem.

[@christopherbradford3778]

Solved it using law of cosines on both smaller triangles and again on the larger triangle. A lot of careful algebra, squaring binomial radical expressions but you eventually get the answer.

[@DamienConcordel]

I did it geometrically myself: Putting A at the origin, I drew the trig circle and placed C' at angle 0 on that circle and E at angle 60° on the same circle. Then I drew a circle with double the radius and placed B' at angle 120° on that bigger circle. This accounts for the bisected angle. I then drew B'C', which intersects AE at D'. Building on known remarkable values for sines and cosines, I figured the coordinates of B' and C' and therefore length B'C': B' = (2 cos 120°, 2 sin 120°) = (-1, sqrt(3)) C' = (1, 0) B'C' = sqrt(2²+sqrt(3)²) = sqrt(7) I then needed to know by what factor to scale the whole diagram to account for AD = 100. This required figuring out AD'. I knew that AE = C'E = 1 (AC'E is equilateral), that AB' = 2, that B'C' and AE intersect at D', and of course that AB' is parallel to C'E. Thales told me that the ratio of AB' to C'E is equal to the ratio of AD' to AE, which led me to the conclusion that AD' = 2/3. So I had to scale the whole diagram up by a factor of 100 / (2/3) = 150. Dilating B' and C' respectively around A by a factor of 150 gave be my actual points B and C, which gave me BC = 150 sqrt(7) (also through Thales / similar triangles)

[@Garygoh884]

If we work out this problem using straightedge and compass, you can see that AB = 3AD using similar triangles. Also, using the Pythagorean theorem, BC = √7 × 3 / 2 AD.

[@pierottialessio]

Used a similar approach. I used the area sine rule. Area ABD = 1/2 *AD*AB*sin(60) = AD*AC*sin(60) whilst Area ACD=1/2*AD*AC*sin(60) (note that it's half of Area ABD because they have same height but ABD base is the double of ACD base, as demonstrated in the video) so Area ABC = Area ABD + Area ACD = 3/2*AD*AC*sin(60). But Area ABC = 1/2*AB*AC*sin(120)=2*(AC^2)*sin(60)*cos(60) so comparing the results and simplifying AC=3/2*AD=150, then AB=300 and using Al-Kashi BC=sqrt[300^2+150^2-2*300*150*cos(120)] which leads to the result

[@mffbataineh]

I have just solved it a a super easy way. From C you draw a line parallel to AD, and from A you extend BA to intersect the the first line in E. This will form an equilateral triangle ACE. We then use similar triangles to find AC, and the law of cosines to find DC. BC is 3(DC): The triangle ACE is equilateral, therefore AE = AC = CE = y, BA = 2y. Using similar triangles: CE/100 = BE/BA=3y/2y, hence y = 150. Using the law of cosines: x^2 = 150^2 + 100^2 - 2(150)(100)cos(60) x = 50 sqrt(7). BC = 150 sqrt(7).

[@titan1235813]

I found it through the following way: trace a perpendicular ED to AB (E on AB), from AB to point D on BC. Same, trace a perpendicular FD to AC (F on AC), from AC to point D on BC. Both perpendiculars have the same length, and they can be found easily from the fact that we know AD's length, and that we have two 90-60-30 right triangles. Now that we have also formed two other right triangles, BED and DCF, we can find their respective hypotenuses, BD and DC, BD in function of BE and DE, and DC in function of CF and DF through the Pythagorean Theorem. Now we find BC in function of AB, AC and cos(120) through the Law of Cosines. That will lead us to the fact that BC is equal to BC√7. We can now see that BC = BD + DC = BC√7. From this last equation we do a few algebraic manipulations, and very easily find that BC = 150√7.

[@iamreal2]

Symbols: △ triangle; ∠ angle; ∟ right angle; ∥ parallel to; ⊥ perpendicular to; ◺right-angled triangle; ≅ congruent to; ~ similar to; ⇒ implies that; ∵ because; ∴ therefore; ° degrees 1) Extend AD to M so that AM = AC = x 2) Connect B to M and extend to N so that CN ∥ AD 3) △AMC is equilateral 4) ∵ AM = x, AB = 2x and ∠BAM = 60° ∴ △BAM is ◺ 5) △BDM is ◺ and CN ∥ AD ⇒ △BCN is ◺ and △BDM ~ △BCN 6) AB = 2AC ⇒ BM : MN = 2 : 1 ∴ BM : BN = 2 : 3 7) ∠CMN = 90° - 60° = 30° ∴ CN = ½x 8) DM = x - AD = x - 100 9) CN : DM = ½x : (x - 100) = 3 : 2 ⇒ x = 150 10) From Gougu in △BNC (or cosine law in △ABC), BC = 150√7

[@КонстантинПосошнов]

Sabc = Sabd + Sadc => 0.5*2x*x*sin(120) = 0.5*2x*100*sin(60) + 0.5*x*sin(60) => x = 150. further by the cosine theorem we find ВС

[@yuanzhilee6405]

Did anyone solve it by this way: First extend line AC to double its initial length, draw a line perpendicular to this extended line before drawing another line to form a right angled triangle. Label the endpoint E. Subsequently, extend line BA by 100 unit lengths , label the new endpoint F. Draw a line from point D which is perpendicular line AC to point F. Label the intersection as G Now notice that both triangles CDG and ABE have angles 30, 60 and 90, respectively. From here, one can deduce that the length of line CE is 2AC * sin(60) = sqrt(3) * (AC) . Further, by Pythagoras theorem , we can calculate the length of line BC which is sqrt(7)*(AC). As the length of AG is 100*cos(60)=50, the length of CG is AC-50. Since triangles CGD and CEB are similar to each other, CE/EB = CG/GD hence ((AB-50)/(50*sqrt(3)))= ((2)/(sqrt(3))) which proves that AB = 150. As such BC = sqrt(7)*(AB) =150*sqrt(7) .

[@jiachengxue913]

You can also extend AD to O and let AO=AB=OB(angle BAD=60). Connect OC you will find OCAB is right angle trapezoid, since cos(DAC)=AC/OA=1/2. In this case, BC=sqrt(OC^2+OB^2)=sqrt(7)*AC. Since AC/OB=AD/OD, then OD=2*AD=200 .So OA=OD+AD=300. Since OA=2AC, AC=150. Thus BC=150*sqrt(7). Your channel remind me the memory of studying in China.

[@nemoumbra0]

I've solved it using the formula for angle bisector's length (AD^2=AB*AC-BD*DC), which could be proved without Stewart's theorem, and the angle bisector's theorem, which we study in Russia a year before the law of sines (AB/AC=BD/DC). I came to 5000=AC^2-CD^2 and next I used cosine theorem for triangle ADC; and that gave me an expression for CD^2. After solving the linear equation for AC I calcutated the CD and, therefore, BC. 150*sqrt(7)

[@benjaminleis979]

Let AD = k for simplicity, 2y = BD and y = DC 1. Extend AD to point E so that ACE is an equilateral triangle. CE || AB and CDE ~ BDA in a 2:1 ratio so AD = 2 DE or DE = 1/2k , therefore AC = AD + DE = 3/2k. 2. Then note AED is a 30-60-90 by SAS since AB = 2AE. There BE = sqrt(3) * AE = 3/2 sqrt(3) k and you can then solve for BD by the Pythagorean theorem and multiple by 3/2 to get all of BC. 3. But its just simpler to instead note from the angle bisector: AB . AC - BD . DC = AD^2 and solve 3/2 k . 3k - 2y^2 = k^2 => y = sqrt(7)/2 . k and in total BC = 3/2 sqrt(7) k

[@bioweapon0073]

Man, all this fancy trigonometry and geometry and i'm here trying to do it graphically. Works just as well, just a wee bit slower. Rotate the shape a little counterclockwise and set A as (0,0), D as (0,-100), and the length of AC as a and AB as 2a. A little trig get us the coordinates of B as (-2asin60, -2acos60) and C as (asin60, -acos60). BDC are colinear so their slopes are the same. Find the slope of BD and DC. Set these two slopes equal to each other and solve for a. Then use the distance formula to find how long BC is.

[@yurenchu]

Yes, that's also how I solved it, except I rotated the shape a bit more, such that D = (100, 0) . I think it's actually faster than Presh's method. There is no need to find the slope of BD: C = (a/2, (a/2)√3), B = (a, -a√3) Since BC is a straight line, and the horizontal x-axis lies at 1/3 of the vertical distance between C and B, that means the x-coordinate of D must lie at 1/3 of the horizontal distance between C and B: D = (a/2 + a/6, 0) = (2a/3, 0) = (100, 0) ==> 2a/3 = 100, a = 150 |BC| = |C-B| = |(-a/2, (3a/2)√3)| = (a/2)*|(-1, 3√3)| = (a/2)*√(1 + 27) = (a/2)*√28 = a√7 = 150√7

[@henryyu316]

At 2:26, angle bisector theorem directly gives you that.

[@actionxp]

extend AD to make an equilateral triangle AEC. you will then be able to a right triangle ABE. then everything will be easy from then.

[@Wildabandon]

I get the answer use Angle bisector theorem, and menelaus. Let C' be a midpoint of AB, then note that C' is reflection C across AD.

[@hbarudi]

What makes it a tricky problem is that you have to use both the law of sines and the law of cosines and some other mathematical rearrangements to solve it such as setting up law of sines over the other to get the needed relationship to solve the problem where most students on a restricted time exam will miss thinking about this situation.

[@davids.9789]

Here's how i got it: First we prove BD=2DC (same way) even tho stewart kills it. Let Q be a point on the segment AB such that AQ=AC now its easy to see AC is perp to AD (30-60-90) and now we'll prove AS=75. S is where CQ and AD meet. Let P be a point on BD such that BP=PD . Now since BQ=AQ and BP=PD It follows that QP=50 and since CS=SQ and CD=DP =>SD=25 thus AS=75 from here we know sc =√3*75 (30-60-90) the rest is phy theorem.

[@IshaaqNewton]

Never stop making videos. Please. I really feel satisfied watching these videos.

[@randomdude9135]

Your name n prof pic lmao

[@tonitalas1757]

Beautiful problem. Excellent explanation!

[@pratyaksh1729]

I tried this problem with the help of law of cosines in multiple triangles. First in ABD to get the value of BD and then in ADC to get the value of DC. Then, I add these two to get the value of BC in terms of x. Next, I get the value of BC in ABC and equating the equations I got a biquadratic equation which I solved using Ferrari's method I got the value of x. Next I put this in the earlier equation to get the value of BC

[@ayushranjan6116]

i solved it in 5 mins😅 using angle bisector theorem and cosine formula EDIT- you also solved the same way 😂😂😂

[@ayushranjan6116]

@@kiwi2539 so what?

[@prod_EYES]

You could use midpoint theorem, 30-60-90 triangle, angle bisector theorem, similar triangles and pretty much solve the whole thing without much trigonometry

[@kooltyme]

I remember doing this problem like a year ago, having a ton of trouble. Now I came back to it and solved it in like 2 minutes it's so easy. I solved it differently, I made BC a y = mx + b equation. Did the same with AB and AC, found the intersections. Distance formula.

[@magdamankowska477]

I did it a bit simpler. I used the area of the triangles ABD, ADC and ABC to figure out x and 2x (using formula with sine). Then I used law of cosines to find BC straight away. I've never heard of Tusi. It though can be done another way.

[@swapnilgupta4256]

Best way to explain

[@Antony_V]

Another bisector theorem: AD=2*[(AC*AB/(AC+AB)]*cos(60°) --> AC=150, AB=300. The rest is the same.

[@marcuscrassus9384]

I've just found another solution where we don't need to know bissector theorem. I did this extending the line BA to E to find a similar triangle that is BCE .It help us with the following formula: BC=[AC*sqrt(400+2AC)]/10

[@swatitripathi9663]

Well, I thought that we could'nt use Trigonometry as In the thumbnail it's written "Geometry Problem".

[@justpaulo]

I just had to use the law of cosines. I used a much better trick to get X. 1) extend down AD till its length is X. Let's call the end point E. 2) △AEC has a 2 sides equal o X and the angle between them is 60°, therefore △AEC is an equilateral triangle. Therefore EC = X. 3) ∠ADB = ∠EDC and ∠BAD = ∠DEC = 60° , therefore △ABD and △ECD are similar triangles (AAA rule). 4) B/c △ABD and △ECD are similar triangles one can say AB/AD = EC/ED ⇒ 2X/100 = X/(X-100) ⇒ X=150 5) The rest is using the law of cosines like in the video.

[@CosmicLegion.]

One of the best you tube channel. Thanks a ton.

[@mffbataineh]

This is my solution to the same problem - I believe it is nicer than the one presented. kzbin.info/www/bejne/omWWYmCBqN1jqpI

[@franciscotrigo9656]

These geometry problems often allow multiples ways to solve them. For example, I find easier to find that BC = 3*DC by mirroring the ADC triangle to the left on the axis AD and inferring that way that the area of ADC is one third of the area of ABC. It would avoid the use of the sine theorem, leaving room for more geometry and less trigonometry. Just a matter of taste! For the second section I found no alternative.

[@latestdancesongsforever7962]

It has other way too Angle opposite to sides are also in the same ratio Angle B= 20° and angle C= 40° Therefore in ∆ADC angle D = 80° Now draw a perpendicular from vertexA on side BC let's say it AE we can calculate AE=sin80°×AD= 100 sine 80° Now in rt∆ AEB and in rt∆AEC calculate BE+EC= AE tan 20°+AE tan40° BC=AE(tan 20°+ tan 40°) BC= 100 sin 80°(tan 20°+ tan40°) Please check the answer