Mathematical and Physical Journal
for High Schools
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Problem C. 1112. (February 2012)

C. 1112. The parallels drawn through a point P of side AB of a triangle ABC to the other two sides intersect the sides AC and BC at the points Q and R, respectively. For what position of the point P is the area of quadrilateral CQPR a maximum?

Suggested by G. Holló, Budapest

(5 pont)

Deadline expired on March 12, 2012.

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

Megoldás. Ossza fel a \(\displaystyle P\) pont az \(\displaystyle AB\) oldalt \(\displaystyle x:1-x\) arányban. \(\displaystyle t_{PRCQ}=t_{ABC}-t_{APQ}-t_{BRP}\). Az \(\displaystyle APQ\) és \(\displaystyle BRP\) háromszögek hasonlóak az eredeti \(\displaystyle ABC\) háromszöghöz, a hasonlóságok aránya \(\displaystyle x\) és \(\displaystyle 1-x\): így az eredeti háromszög \(\displaystyle T\) területével kifejezhető a \(\displaystyle CQPR\) paralelogramma területe: \(\displaystyle T-x^2 T - (1-x)^2 T=(2x^2-2x)T=2\left((x-\frac{1}{2})^2-\frac{1}{4}\right)T\). Tehát a \(\displaystyle CQPR\) négyszög területe akkor a legnagyobb, amikor \(\displaystyle P\) az \(\displaystyle AB\) oldal felezőpontja: ekkor a négyszög területe fele a háromszögének.


221 students sent a solution.
5 points:160 students.
4 points:17 students.
3 points:10 students.
2 points:8 students.
1 point:15 students.
0 point:9 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, February 2012