ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
41 lines
1.3 KiB
Markdown
41 lines
1.3 KiB
Markdown
---
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id: 5900f3fc1000cf542c50ff0e
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title: 问题143：研究三角形的Torricelli点
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challengeType: 5
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videoUrl: ''
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dashedName: problem-143-investigating-the-torricelli-point-of-a-triangle
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---
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# --description--
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设ABC为三角形，所有内角均小于120度。设X为三角形内的任意点，并使XA = p，XC = q，XB = r。 Fermat挑战Torricelli找到X的位置，使p + q + r最小化。 Torricelli能够证明，如果在三角形ABC的每一侧构造等边三角形AOB，BNC和AMC，则AOB，BNC和AMC的外接圆将在三角形内的单个点T处相交。此外，他证明了T，称为Torricelli / Fermat点，最小化p + q + r。更值得注意的是，可以证明，当总和最小化时，AN = BM = CO = p + q + r，并且AN，BM和CO也在T处相交。
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如果总和最小化并且a，b，c，p，q和r都是正整数，我们将称三角形ABC为Torricelli三角形。例如，a = 399，b = 455，c = 511是Torricelli三角形的示例，其中p + q + r = 784.找到Torricelli三角形的p + q +r≤120000的所有不同值的总和。
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# --hints--
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`euler143()`应返回30758397。
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```js
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assert.strictEqual(euler143(), 30758397);
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```
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# --seed--
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## --seed-contents--
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```js
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function euler143() {
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return true;
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}
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euler143();
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```
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# --solutions--
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```js
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// solution required
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```
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