Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem B. 4344. (March 2011)

B. 4344. The lengths of the parallel sides of a symmetric trapezium are a and c. The midpoints of the legs are E and F. Let G denote the orthogonal projection of point E on the line of the leg BC. What is the area of the trapezium if point C divides the line segment GF in a 1:2 ratio? (C lies closer to point G.)

(4 pont)

Deadline expired on April 11, 2011.

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

Megoldás. A feladat szövege alapján világos, hogy az \(\displaystyle E\) és \(\displaystyle F\) pontok rendre az \(\displaystyle AD\), illetve a \(\displaystyle BC\) szár felezőpontjai, továbbá az \(\displaystyle AB\) alap hosszabb mint a \(\displaystyle CD\) alap. Ezek szerint \(\displaystyle a\ne c\). Tegyük fel, hogy \(\displaystyle a>c\). A szárak hossza legyen \(\displaystyle 4x\), ekkor \(\displaystyle FG=3x\). Ha a \(\displaystyle C\) csúcsnak az \(\displaystyle AB\) alapra eső merőleges vetülete \(\displaystyle H\), akkor \(\displaystyle EF=\frac{a+c}{2}\) és \(\displaystyle BH=\frac{a-c}{2}\). Az \(\displaystyle EFG\) és \(\displaystyle CBH\) derékszögű háromszögek hasonló volta miatt

\(\displaystyle 12x^2=CB\cdot FG=EF\cdot BH=\frac{a+c}{2}\cdot \frac{a-c}{2} =\frac{a^2-c^2}{4}.\)

Ezek alapján a trapéz területe

\(\displaystyle EF\cdot CH=EF\cdot\sqrt{CB^2-BH^2}= \frac{a+c}{2}\cdot \sqrt{(4x)^2-\left(\frac{a-c}{2}\right)^2}\)

\(\displaystyle =\frac{a+c}{2}\cdot \sqrt{\frac{a^2-c^2}{3}-\left(\frac{a-c}{2}\right)^2} =\frac{a+c}{2}\cdot \sqrt{\frac{a^2+6ac-7c^2}{12}}\)

\(\displaystyle =\sqrt{\frac{(a+c)^2(a+7c)(a-c)}{48}}.\)


87 students sent a solution.
4 points:62 students.
3 points:8 students.
2 points:3 students.
1 point:7 students.
0 point:5 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, March 2011