a unique solution for finding red line length? | (Fun Geometry Problem) |

[@matthieudutriaux]

BC=BS+SC BC=x/tan(Pi-2*b) ; BS=x/tan(a) ; SC=TC=6 BC=x/tan(Pi-2*b)=x/tan(-2*b)=-x/tan(2*b) Same calculus as you, Math and Engineering to find : tan(b)=4*sqrt(2)/2=2*sqrt(2) (5:08) And to find tan(a) since sin(a)=sqrt(6)/3 and cos(a)=sqrt(3)/3 (10:16) Then, tan(2*b)=2*tan(b)/(1-(tan(b))^2)=2*2*sqrt(2)/(1-8)=-4*sqrt(2)/7 And tan(a)=sin(a)/cos(a)=sqrt(2) BC=BS+SC -x/tan(2*b)=x/tan(a)+6 7*x/(4*sqrt(2))=x/sqrt(2)+6 7*x/4=x+6*sqrt(2) 7*x/4-x=6*sqrt(2) 3*x/4=6*sqrt(2) x=4/3*6*sqrt(2) then x=8*sqrt(2)

[@MathandEngineering]

Oh wow, this method is really smart and interesting, using tan even tan(b) made it more fast and yes the final answer is accurate, thanks, I love the method.

[@ducduypham7264]

In isosceles triangle STC with base ST, SA is external angle bisector so ST/SC=AT/AC or 4/6=AT/(AT+6). Solve for AT we have AT=12. In isosceles triangle STC, draw a perpendicular from C to ST and intersect ST at M. We can easily prove MT=2. Draw a perpendicular from A to ST that intersect ST at N. We can easily prove that triangle CMT similar to triangle ANT with similar ratio equal 1/2 (as CT/AT=6/12) then TN=2*TM=2*2=4. Right triangle ATM with AT=12 is hypotenuse, TN=4 so we can conclude that AN=8*sqrt(2). Triangle ABS congruent to triangle ANS then x=AB=AN=8*sqrt(2)

[@MathandEngineering]

Wow I totally forgot about the exterior angle bisector theorem when I was solving the Question, method is creative. I wish students can think as critical as this, one of the best things that can happen to a man is the ability of thinking out of the box and that is what you did here, it's great

[@christiannehman7846]

Interesting Problem and Great Solution ! Thank you for Sharing . I found a purely Geomtric Solution (No Trigonometry involved) that I would like to share . -'First Step : New points Start by drawing the feet of the perpendicular from A to (ST) , let's call it D. And let's call E the one from C to (ST). -Second Step: Similar triangles ΔATD ≡ ΔCTE ( ≡ means similar): Let k be the ration of ATD to CTE : Then : AT=TC.k=6k TD=ET.k=2k AD=EC.k=4sqrt(2)k -Third Step : Congruent Triangles Notice that ΔABS=ΔADS Therefore: AB=AD=4sqrt(2).k BS=SD=ST+TD=4+2k Last Step : Pythagoras Theorem ABC is a right triangle ,thus: AB²+BC²=AC² (4sqrt(2).k)²+(2k+10)²=(6k+6)² ... k=2 Therefore AB=4sqrt(2).k AB=8sqrt(2) QED

[@MathandEngineering]

Wow this is really creative, it is time saving, and also gives and accurate answer, I see no room for error in the method. Thanks for sharing, it is really nice

[@christiannehman7846]

Your welcome , all my appreciation for your work . Have a nice day!

[@mohabatkhanmalak1161]

I got lost somewhere in the middle there when it got into the serious trigonometry. But I could still follow the path, only I have forgotten those trig identities. Thanks for sharing, its wonderfull.☘

[@MathandEngineering]

Thank you, it's my pleasure that you like the video,

[@Mipullo321]

Thanks, i enjoyed watching the video, the way you did the solving is really professional, i cant wait to start solving math like this

[@MathandEngineering]

Yes mr/Mrs mipullo, I believe in you, you can do very much better than this, just put more effort and patience, you'll discover the greatness in you. Have a good day

[@Jahaaanaa]

I have tried to use other methods to solve it, but without calculator, its impossible for me

[@MathandEngineering]

No it's not impossible for you, you are a great person, don't let this little maths Question make you underrate yourself, there are other way you can get there without using calculator, just don't give up, pay attention to the Question, give it sometime I am sure that you'll find the answer, try every possible means you can think of