Equiangular Hexagon

Geometry Level 5

Equiangular hexagon $ABCDEF$ has $AB=BC=DE=EF=4$ and $AF=CD=1$ . The hexagon is reflected across the segment $EF$ to form hexagon $A'B'C'D'EF$ . $\overline{B'C}$ intersects $\overline{EF}$ at point $P$ . $\dfrac{EP}{FP}$ can be expressed as $\dfrac{p}{q}$ for relatively prime positive integers $p,q$ . Find $p\times q$ .

Inspired by a friend.

The answer is 7.

4 solutions

Lawrence Pauls
Oct 29, 2016

Extend AB and EF so they intersect at G. Triangle AGF is an equilateral triangle so GF = 1
Draw segment BB' and label the point Q where it intersects segment EF.
Triangle BGQ is a 30-60-90 right triangle with BG = 5 therefore GQ = 2.5 and FQ = 1.5.
Triangle B'BC is similar to B'QP and is twice the size, therefore QP is half of BC which is 2.
FP = FQ + QP = 1.5 + 2 = 3.5
EP = 4 - 3.5 = 0.5
the ratio is 7 to 1 so the answer is 7

In the 4th step, it will be-"......PQ(not FQ) is half of....".

ARUNEEK BISWAS - 4 years, 7 months ago

Yes, thanks I fixed the typo.

Lawrence Pauls - 4 years, 7 months ago

Khondaker Sadman
Apr 21, 2014

This problem can be solved pretty easily by coordinate-bash since the angles are so nice. If we have the coordinates of F, P, and E then we can easily find lengths EP and FP. Defining F to be our origin (0,0), E is readily seen to be (4,0) since FE is length 4. Then we just need to determine P. To determine P, note that it can be considered the x-intercept of the line B'C. So if we can get an equation for the line B'C, we can find the x-intercept and that would be P.
To get an equation for the line B'C, we need the coordinates of 2 points, naturally the 2 points that we'll find coordinates for are B' and C. We can get the coordinates of C by starting at E, then traversing the hexagon up to D and then going up to C. Noting that the acute angle ED makes witht the horizontal is 60 degrees, we can draw a right triangle with hypotenuse 4 (ED) to see that point D is 4 sin(60) higher than E and 4 cos(60) farther to the right than E. This means that D is (4+4cos(60), 0+4sin(60)) = (6, 2sqrt(3)). Then we can get point C by noting that the angle DC makes with the vertical is 30 degrees, we can draw a right triangle with hypotenuse 1 (CD) to see that point C is sin(30) to the left of D and cos(30) up from D. This means taht C is (6-sin(60), 2sqrt(3)+cos(30)) = (5.5, 2.5sqrt(3)).
We can follow an identical process to determine point B' (or just note that the vector FB' is just the vector EC' reflected over a vertical axis). Then we'll have B' is (1.5,-2.5sqrt(3)).
Now that we have both points, we can find the slope and then write the point-slope form of the line, which ends up being y-2.5sqrt(3) = 5sqrt(3)/4 (x-5.5). Then solve the y-intercept (P) to be (3.5,0). At this point, it's straightforward to determine EP and FP.

Tanishq Kancharla
Apr 6, 2014

To do this, I drew a diagram. After, I drew where lines BC and ED intersect (and labeled this G). Then I noticed that BGE was an equilateral triangle. In fact, drawing the intersections of BA and EF gives us a rhombus with length 5 instead of a hexagon. This is easier to deal with because we notice that B'PE is similar to B'CG. This is because angles B'GC=60 degrees=BEP because they're all equilateral triangles. Now the ratio of sides is simply 1/2 because B'G=BE+EG=5+5=10 while B'E=5. So now PE=CG/2=1/2. Note that CG is 1 because CD is an equilateral triangle with side length 1. Now that PE=1/2, we have FP=4-1/2=7/2 and PE/FP=1/7 which means pq=1*7=7

Dinesh Inspire
Mar 25, 2014

The Key thing here is the prime factors where the ratio between them should in the form of p/q where p and q are primes . EP/FP = 0.5/3.5=1/7 and p x q = 7 .

how is that the key thing?

Tanishq kancharla - 7 years, 2 months ago

Equiangular Hexagon

The answer is 7.

4 solutions

0 pending reports