ACTUALLY IT WAS PRETTY EASY!!! PICK AN ARBITARY POINT ON UPPER LINE.NAME IT 'A'. THEN SEGREGATE A 30 DEGREE ANGLE THAT CUTS THE LOWER LINE AT POINT 'B'. LINE BISECTION of 'AB' MEETS THE THIRD LINE WHICH IS THE BETWEEN 2 ELSE ONES. NAME THE LAST POINT 'O'. CONSEQUENTLY IF ONE YOU DRAW A CIRCLE WITH CENTER 'O' AND ALSO LENGTH 'OA' AS RADII; YOU CAN GET YOUR POPOSE, AN EQUILATERAL TRIANGLE.THE ONLY PROBLEM YOU MAYBE STUCK WITH IS HOW TO GET A 30 DEGREE ANGLE? THAT'S ALSO SUCH AN EASE. YOU GOTTA CONSTRUCT AN EQUILATERAL TRIANGLE. NEXT YOU SHOULD BISECT A N ANGLE AS U WISH TO CHOOSE. AFTERWARDS U JUST NEED TO BUILD 2 TRIANGLE CONTAINING THAT 30 ANGLE TOO ON BOTH SIDES, ARBITARY EQUILATERAL TRIANGLE & POINT 'A' WHICH CONCLUDE UPPER LINE AS AN SIDE OF TRIANLE AND SO ON. PROVING THE CONSTRUCTURE ABOVE IIS ON READER. ITS SO STRAIGHT LIGHT.

Another solution: Take a perpendicular which crosses top and middle lines in A and B resp. Build equilateral triangle ABC. Orthogonal projection of C on bottom line is P. Take O as middle of AB. Perpendicular to OP in O crosses top and middle lines in Q and R. PQR is equilateral.

Let's name the paralleles L1, L2, L3 (L2: middle) and A and B the distances among L1,L2 and L2,L3. The side S of the searched triangle is S=sqrt(A²+B²+AB)/(sqrt(3)/2)

If we build a triangle PQR with PQ=A, QR=B and <PQR=120º, then PR=q=sqrt(A²+B²+AB). Build now an equilateral triangle PRZ. The altitudes (heights) of PRZ are q/(sqrt(3)/2)=sqrt(A²+B²+AB)/(sqrt(3)/2)=S.

Sorry, my mistake in last post. 1) Draw a perpendicular crossing paralleles in A,B,C 2) Draw equlateral triangle BCD 3) Draw triangle ADE, with <ADE=30º and <DAE=90º

César, Your answer is correct. Long ago, I was planning to ask the general case, you've noted. See the last one on page: http://www.8foxes.com/Home/brain-teaser

That may be one of the first foxes ever constructed (That's why it is not numbered). But upon your solution, I will re-design and post it in the following days. Symmetry in the formula deserves it!

Another solution. Take a point Pon the lowest line and erect PX at 60 to it meeting the second line at x. draw angle XPY=60 meeting the third line at Y. draw angle XYZ=60 meeting the first line at Z. Then XYZ is the required triangle. proof: YZPX is cyclic as external angle=int opp angle, so YZX=YPX=60 Murali

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ACTUALLY IT WAS PRETTY EASY!!!

ReplyDeletePICK AN ARBITARY POINT ON UPPER LINE.NAME IT 'A'. THEN SEGREGATE A 30 DEGREE ANGLE THAT CUTS THE LOWER LINE AT POINT 'B'. LINE BISECTION of 'AB' MEETS THE THIRD LINE WHICH IS THE BETWEEN 2 ELSE ONES. NAME THE LAST POINT 'O'. CONSEQUENTLY IF ONE YOU DRAW A CIRCLE WITH CENTER 'O' AND ALSO LENGTH 'OA' AS RADII; YOU CAN GET YOUR POPOSE, AN EQUILATERAL TRIANGLE.THE ONLY PROBLEM YOU MAYBE STUCK WITH IS HOW TO GET A 30 DEGREE ANGLE? THAT'S ALSO SUCH AN EASE. YOU GOTTA CONSTRUCT AN EQUILATERAL TRIANGLE. NEXT YOU SHOULD BISECT A N ANGLE AS U WISH TO CHOOSE. AFTERWARDS U JUST NEED TO BUILD 2 TRIANGLE CONTAINING THAT 30 ANGLE TOO ON BOTH SIDES, ARBITARY EQUILATERAL TRIANGLE & POINT 'A' WHICH CONCLUDE UPPER LINE AS AN SIDE OF TRIANLE AND SO ON. PROVING THE CONSTRUCTURE ABOVE IIS ON READER. ITS SO STRAIGHT LIGHT.

GOOD LU(o)CKS, UNLOCK IT PLEASE!

Another solution:

ReplyDeleteTake a perpendicular which crosses top and middle lines in A and B resp. Build equilateral triangle ABC. Orthogonal projection of C on bottom line is P. Take O as middle of AB. Perpendicular to OP in O crosses top and middle lines in Q and R. PQR is equilateral.

bleaug

Let's name the paralleles L1, L2, L3 (L2: middle) and A and B the distances among L1,L2 and L2,L3. The side S of the searched triangle is

ReplyDeleteS=sqrt(A²+B²+AB)/(sqrt(3)/2)

If we build a triangle PQR with PQ=A, QR=B and <PQR=120º, then PR=q=sqrt(A²+B²+AB). Build now an equilateral triangle PRZ. The altitudes (heights) of PRZ are

q/(sqrt(3)/2)=sqrt(A²+B²+AB)/(sqrt(3)/2)=S.

César Lozada, Venezuela

Sorry, my mistake in last post.

ReplyDelete1) Draw a perpendicular crossing paralleles in A,B,C

2) Draw equlateral triangle BCD

3) Draw triangle ADE, with <ADE=30º and <DAE=90º

DE is the side of the target triangle.

César Lozada, Venezuela

César,

ReplyDeleteYour answer is correct. Long ago, I was planning to ask the general case, you've noted. See the last one on page:

http://www.8foxes.com/Home/brain-teaser

That may be one of the first foxes ever constructed (That's why it is not numbered). But upon your solution, I will re-design and post it in the following days. Symmetry in the formula deserves it!

Gracias!

Another solution. Take a point Pon the lowest line and erect PX at 60 to it meeting the second line at x. draw angle XPY=60 meeting the third line at Y. draw angle XYZ=60 meeting the first line at Z. Then XYZ is the required triangle. proof: YZPX is cyclic as external angle=int opp angle, so YZX=YPX=60

ReplyDeleteMurali

Do not use all of these Private Money Lender here.They are located in Nigeria, Ghana Turkey, France and Israel.My name is Mrs.Ramirez Cecilia, I am from Philippines. Have you been looking for a loan?Do you need an urgent personal or business loan?contact Fast Legitimate Loan Approval he help me with a loan of $78.000 some days ago after been scammed of $19,000 from a woman claiming to be a loan lender from Nigeria but i thank God today that i got my loan worth $78.000.Feel free to contact the company for a genuine financial call/whats-App Contact Number +918929509036 Email:(fastloanoffer34@gmail.com)

ReplyDelete