Mathematical and Physical Journal
for High Schools
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Problem B. 5114. (September 2020)

B. 5114. A unit cube \(\displaystyle ABCDEFGH\) (see the figure) is cut by a plane \(\displaystyle \mathcal{S}\) that intersects the edges \(\displaystyle AB\) and \(\displaystyle AD\) at the points \(\displaystyle P\) and \(\displaystyle Q\) respectively, such that \(\displaystyle AP=AQ=x\) (\(\displaystyle 0<x<1\)). Let the common point of the edge \(\displaystyle BF\) and \(\displaystyle \mathcal{S}\) be \(\displaystyle R\). What is the distance \(\displaystyle BR\) if \(\displaystyle \angle QPR =120^\circ\)?

(4 pont)

Deadline expired on October 12, 2020.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Legyen \(\displaystyle AP=AQ=x\) és \(\displaystyle BR=y\). Feltehető, hogy \(\displaystyle P\) az \(\displaystyle AB\) él belső pontja, azaz \(\displaystyle 0<x<1\), különben \(\displaystyle P=Q\) vagy \(\displaystyle P=R\), és a \(\displaystyle QPR\angle\) nem jön létre.

A Pitagorasz-tétel többszöri alkalmazásával kapjuk, hogy \(\displaystyle PQ^2=AQ^2+AP^2=2x^2\), \(\displaystyle PR^2=BP^2+BR^2=(1-x)^2+y^2\) és \(\displaystyle QR^2=AQ^2+AB^2+BR^2=x^2+y^2+1\).

Írjuk fel a koszinusz-tételt a \(\displaystyle PQR\) háromszögben a \(\displaystyle P\angle\)-re: \(\displaystyle QR^2=PQ^2+PR^2-2 \cdot PQ\cdot PR \cdot \cos 120^\circ\). Ebbe beírva a fentieket kapjuk, hogy

\(\displaystyle x^2+y^2+1=2x^2+(1-x)^2+y^2-2\cdot \sqrt{2x^2} \cdot \sqrt{(1-x)^2+y^2}\cdot \frac {-1}2.\)

Rendezés majd négyzetreemelés után:

\(\displaystyle (-2x^2+2x)^2=(2x^2)\cdot ((1-x)^2+y^2).\)

Kihasználva, hogy \(\displaystyle x>0\), oszhatunk \(\displaystyle 2x^2\)-tel, így kapjuk, hogy \(\displaystyle 2(x-1)^2=(x-1)^2+y^2\), azaz \(\displaystyle y^2=(x-1)^2\). Ebből \(\displaystyle x< 1\) és \(\displaystyle y>0\) miatt \(\displaystyle y=1-x\) következik, azaz a \(\displaystyle BR\) távolság \(\displaystyle 1-x\).


Statistics:

106 students sent a solution.
4 points:84 students.
3 points:10 students.
2 points:3 students.
1 point:3 students.
0 point:1 student.
Unfair, not evaluated:5 solutionss.

Problems in Mathematics of KöMaL, September 2020