<< problem 581 - 47-smooth triangular numbers Divisibility streaks - problem 601 >>

# Problem 587: Concave triangle

A square is drawn around a circle as shown in the diagram below on the left.
We shall call the blue shaded region the L-section.
A line is drawn from the bottom left of the square to the top right as shown in the diagram on the right.
We shall call the orange shaded region a concave triangle.

It should be clear that the concave triangle occupies exactly half of the L-section.

Two circles are placed next to each other horizontally, a rectangle is drawn around both circles, and a line is drawn from the bottom left to the top right as shown in the diagram below.

This time the concave triangle occupies approximately 36.46% of the L-section.

If n circles are placed next to each other horizontally, a rectangle is drawn around the n circles, and a line is drawn from the bottom left to the top right, then it can be shown that the least value of n for which the concave triangle occupies less than 10% of the L-section is n = 15.

What is the least value of n for which the concave triangle occupies less than 0.1% of the L-section?

# My Algorithm

This is a so-called paper'n'pen problem because you can solve it without a computer.
But I like computers so much that I decided to write a proper program that solves this problem by throwing CPU cycles at it !

A little mathematics is still required:

• the line from the lower left corner to the upper right corner has the equation y = mx where m is the slope
• I define that my circles have a radius r = 1
• so that the box has a height of r+r = 2
If there is only one circle then it covers an area of pi r^2 = pi 1^2 = pi. Its bounding box covers an area of 2^2 = 4.
There are four L-shaped areas between circle and box - all have the same size, so that one L-shaped area covers an area of
(1) (4 - pi) / 4 approx 0.2146

To find the intersection between the line and the first circle - which is centered at (1,1) - I solve both equation for y:
(2) y = mx
(3) (x - 1)^2 + (y - 1)^2 = 1
(4) (y - 1)^2 = 1 - (x - 1)^2
(5) y - 1 = \pm sqrt{1 - (x - 1)^2}
(6) y = 1 - sqrt{1 - (x - 1)^2} ← I only need the lower half of the circle where the first intersection takes place

The intersection is at (2) = (6):
(7) mx = 1 - sqrt{1 - (x - 1)^2}

And that's when I decided to write a "true" IT solution and solve equation (7) numerically:
• compute left and right side of equation (7)
• if they are "close enough", then abort
• else move x accordingly
• see getIntersection() (where slope is m)
Knowing the intersection I proceed with the same motivation: keep solving it numerically !
• function getAreaL() computes the area of the L-shaped area given a slope
• left of the intersection point is a triangle with area x * y / 2 = x * m x / 2
• right of the intersection point is the integral of equation (6)
• my numerical integration takes 100000 steps between the intersection point and x = 1
At first I checked every possible size of 1,2,3,...,100,... circles whether the area is below 0.1%.
It took a little more than one second to find the correct solution.
Knowing that the area is strictly decreasing I re-wrote my code to work similar to binary intersection:
check only every 64th number: 64, 128, 192, ... circles and when the area is below the limit then go back and check every 32nd, then 16th, ... until step = 1.
Now the result is printed after less than 0.1 seconds.

## Alternative Approaches

As I told you before, you can solve this with pen'n'paper (or Wolfram Alpha).

## Note

Of course there are a few assumptions which are not always true:

• maybe Epsilon is too large and the result will be wrong
• maybe I need more than 100000 steps to accurately evaluate getAreaL
→ but my default values easily solve the problem

I had fun writing this solution - for whatever reason I like numerical approaches such as Monte-Carlo simulations.

# Interactive test

This feature is not available for the current problem.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <cmath>

const auto Epsilon = 0.00000000001;
const auto NoLine  = 0;

// return x where line and first circle intersection
double getIntersection(double slope)
{
// circle: (x-1)^2 + (y-1)^2 = 1
//                   (y-1)^2 = 1 - (x-1)^2
//                    y-1    = sqrt(1 - (x-1)^2)
//                         y = 1 - sqrt(1 - (x-1)^2)
// line:                   y = mx    <= where m is the slope
// their intersection:    mx = 1 - sqrt(1 - (x-1)^2)

// I entered those formula in Wolfram Alpha and didn't like the output ... so let's solve this numerically !
// choose any x in 0 < x < 1
auto x    = 0.5;
// adjust x by this value if not precise enough
auto step = 0.1;

while (true)
{
// good enough ?
auto leftSide  = slope * x;
auto rightSide = 1 - sqrt(1 - (x - 1)*(x - 1)); // see above how I got this formula
if (fabs(leftSide - rightSide) < Epsilon)
return x;

// move x
if (leftSide > rightSide)
x -= step;
else
x += step;

// take smaller steps
step *= 0.99;
}
}

// return area of the L-shaped area
double getAreaL(double slope)
{
// x-coordinate of first intersection between line and circle
double intersection = 0;
// area of the triangle on the left side of the intersection
double leftArea     = 0;
if (slope > 0)
{
intersection = getIntersection(slope);
leftArea = intersection * (intersection * slope) / 2;
}

// the area on the right side of the intersection can be computed exactly - but I hate ugly looking formulas and prefer C++ code :-)

// numerically integrate area between intersection and "where the first circle touches the box"
auto rightArea = 0.0;
auto Step = (1 - intersection) / 100000;
for (double x = intersection; x < 1; x += Step)
{
double y = 1 - sqrt(1 - (x - 1)*(x - 1)); // same formula as in getIntersection
rightArea += y * Step;
}

return leftArea + rightArea;
}

int main()
{
// stop if below 0.1%
auto limit = 0.1;
std::cin >> limit;

// a circle has an area of A=pi * r^2
// r=1 => A = pi
// the box around the circle has an area of 2^2 = 4
// there are four L's between circle and box, I only need the one in the lower half
// L = (4 - pi) / 4 = 0.21460183660255169038433915418012...
auto L = getAreaL(NoLine);   // NoLine=0 is a special case in getAreaL

auto numCircles = 1;
auto step = 64; // powers of two are great, but could be any other positive number, too (even 1)
while (true)
{
auto slope = 1.0 / numCircles;
auto area  = getAreaL(slope);
// area small enough ?
auto percentage = 100 * area / L;
if (percentage < limit)
{
// reached small step size, found solution
if (step == 1)
break;

// maybe I "jumped over" the correct solution, go back and reduce step size
numCircles -= step;
step /= 2;
}

// look at next slope
numCircles += step;
}

// and print the solution
std::cout << numCircles << std::endl;
return 0;
}


This solution contains 16 empty lines, 29 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.04 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

October 23, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300
 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400
 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500
 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638
The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 581 - 47-smooth triangular numbers Divisibility streaks - problem 601 >>
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !