Mathematical and Physical Journal
for High Schools
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Problem C. 1077. (April 2011)

C. 1077. E is the point dividing side AC of triangle ABC in a 3:1 ratio, such that EC is the shorter part. The line passing through E and the midpoint F of side BC intersects the line AB at D. What percentage is the area of triangle ADE of the triangle ABC?

(5 pont)

Deadline expired on May 10, 2011.

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

Megoldás. A háromszög \(\displaystyle B\) csúcsából húzzunk párhuzamost az \(\displaystyle AC\) oldallal: ez \(\displaystyle G\)-ben metszi \(\displaystyle ED\)-t. \(\displaystyle FBG_\triangle \cong FCE_\triangle\), mert a párhuzamos oldalaik miatt hasonlóak és \(\displaystyle F\) felezőpont miatt \(\displaystyle CF=FB\) (azaz \(\displaystyle F\)-re középpontosan tükrösek), így \(\displaystyle BG=EC=AC/4=x\) (és \(\displaystyle AE=3x\)). \(\displaystyle ADE_\triangle \sim BDG_\triangle\) az egyállású szögeik miatt, így az oldalakra \(\displaystyle \frac{x}{3x}=\frac{GB}{EA}=\frac{BD}{AD}=\frac{y}{3y}\), azaz \(\displaystyle AB=2y\). A területek aránya pedig

\(\displaystyle \frac{t_{ADE}}{t_{ABC}}=\frac{{1 \over 2}AE\cdot AD\cdot \sin\alpha}{{1 \over 2}AC\cdot AB\cdot \sin\alpha}=\frac{9xy}{8xy}=\frac 98 .\)

Az \(\displaystyle ADE\) háromszög területe \(\displaystyle 112,5\%\)-a az \(\displaystyle ABC\) háromszög területének.


117 students sent a solution.
5 points:103 students.
4 points:7 students.
3 points:2 students.
2 points:1 student.
0 point:4 students.

Problems in Mathematics of KöMaL, April 2011