As you might have noticed, Sunday Chess Problem had the week off. If you really need to get your fix, though, you can have a look at this web page I made for my chess problems. You'll recognize a few of them from the Sunday Chess Problem series.

I did, however, manage to get the new POTW up. Number seven, this week. I also posted an “official” solution to last week's problem. So go have a look and let me know what you think!

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Nice problem. Your intro gives a subtle hint. Aside from the aha!s it is quick to get 24sqrt5.

What Poincare said.

Having the equal perimeters allows you to solve for the length CD, which comes out 60. Dropping a vertical from D down to BC (let the intersection at BC be point E) gives you a right triangle DEC similar to (at 60% scale) ABC. The vertical DE is therefore 48 units, EC is 36 units, and so BE is 24 units. BED is also a right triangle, and with the two legs known, the hypotenuse is 24*sqrt(5).

Assuming my logic here is correct, I found this a lot easier than the last one.

Pythagorus tells us that AC = 100.

The problem statement is that AB + AD + BD = BC + BD + CD

or that 80 + AD + BD = 60 + BD + CD

ie AD - CD = -20

and AD + CD = 100

ie 2AD = 80

ie AD = 40, CD = 60

With CD = 60, that means that triangle BCD is isosceles. And since it was a right angled triangle, we know that cos C = 0.6. If we choose a point E on BD such that CE bisects C, it will be a right angled triangle (as BCD is isosceles). DE/CD = sin C/2, so BD = 60 * sin C/2 * 2 which works out to about 53.666.

I confess I'm not entirely clear on how I would solve this had it not turned out that one of the inner triangles was isosceles though. Is it still solvable in such a case?

By the way, 24 * sqrt(5) is indeed 53.666; obviously the two posters above me deserve more credit for an exact answer, but it's gratifying to know that my answer is nonetheless correct (if arrived at a slightly different way).

Michael,

I like your solution. It is far more elegant than mine, and results in a much simpler mathematical expression for the answer. I was able to use the equality of perimeters and the Pythagorean theorem to find pretty easily that AD=40 and DC=60. From there I applied the law of sines, utilizing the fact that triangle BCD turns out to be isosceles. The sine of angle DCB is obviously 0.8. Angle BDC (and DBC) is equal to 1/2*(pi-arcsin(.8)), which I will label c to avoid having to type this expression again. Therefore, sin c/60 = 0.8/DB which rearranges to DB sin c = 48 or DB = 48 / sin c. I checked it out numerically, and my solution does indeed match the solutions already posted.

I know it is intended to be a geomerty problem, but once we know that CD = 60 so triangle BCD is isosceles, we can treat it as a trig problem. The angle at C is arctan(80/60). We know that the base on an isosceles triangle is equal to one of the two equal sides times twice the sin of half the angle between the two equal sides, so BD = 60*2*sin(arctan(80/60)/2), or about 53.666.

I dropped a perpendicular from the apex of the isosceles triangle BCD to BD, solving for the length of half its base BD. Cos angle BDC was required, but I was rather hoping that there might be a non-trigonometric solution, as Jason hasn’t previously required that we invoke trig in his problems. Can anyone tell me if there is such a solution, or is it necessarily a trigonometric problem?

Phil-

Yes, there is a geometric solution without trig. #2 above sets it out.

Michael Kelsey's solution above uses only similar triangles and pythagorean theorem. No trig necessary.

Oh dear. My mathematical ignorance exposed. Can someone explain to me what 24*sqrt(5). actually means? Thanks.

Similar to the solution of Michael Kelsey in #2 above, instead of dropping the perpendicular from D down to BC, drop it horizontally to point E on line AB. The resuting triangle AED is similar to triange ABC with a scale of .4, so line ED = 24 and AE = 32, thus EB = 48. Which means BD = sqrt(24^2 + 48^2) = 24*sqrt(5).

Phil- It means 24 times the square root of 5. This is the way it is written in many (most?) computer languages, which use * as the times sign and sqrt() as the square root function. This usage arose due to the limitations of the character set on earlier computer consoles and has stuck around as a kind of de facto standard.

In another approach, of which Jason will probably not approve of in this context, we can place the triangle ABC on a coordinate frame with B at (0, 0), A at (0, 80) and C at (60, 0). We can then use parameter s, the distance from C in the direction of A, to descrive line AC as x = 60 - .6*s and y = .8*s. Since the length of AD = 60, the coordinates of point D are (24, 48). The distance from point B, at (0,0) and point D at (24, 48) is thus 24*sqrt(5).

Typo: the lenght of CD = 60.

Fun problem, especially since I've not done any trig since high school, many many years ago. But I think I did it a dumb way:

I got to figuring out CD=60 the same way as Gary, then that makes BD the chord of angle at C. Had to look up the equation for length of a chord but was gratified to see it was simple, and got BD=53.666

Thanks Rick. If I'd taken any notice of the sensible move in #2 to create the similar triangles by dropping perpendicular DE, then I guess I could have answered my own question - in which case, with a little applied intelligence, I might even have been able to work out for myself what 24*sqrt(5) could mean! Oh well, we live and learn.

jrosenhouse wrote (March 23, 2015):

>

the new POTW […] Number sevenand there ( http://educ.jmu.edu/~rosenhjd/POTW/Spring15/POTW7S15.pdf )

>

Segment BD is drawn to the hypotenuse [AC] in such a way that triangles ABD andBCD have the same perimeter.

Segment BD is therefore one ot the three so-called splitters of triangle ABC.

Jason, a quibble: it's Desargues' theorem, not Desargue's; that is, the chap in question was called Desargues not Desargue.

g-

Thanks for pointing out the error!