Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem B. 4889. (September 2017)

B. 4889. The trapezium $\displaystyle ABCD$ has an inscribed circle. The circle touches base $\displaystyle AB$ at point $\displaystyle T$, and the parallel base $\displaystyle CD$ at point $\displaystyle U$. Let $\displaystyle M$ denote the intersection of the lines of legs $\displaystyle AD$ and $\displaystyle BC$, and let $\displaystyle V$ be the intersection of side $\displaystyle AB$ with line $\displaystyle MU$. Show that $\displaystyle AT=VB$.

(4 pont)

Deadline expired on October 10, 2017.

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

Megoldásvázlat. Legyen $\displaystyle AV=x$, $\displaystyle VT=z$, $\displaystyle TB=BE=y$, $\displaystyle FD=DU=t$, $\displaystyle UC=CE=s$, $\displaystyle AF=AT=x+z$. A beírt kör átmérőjének négyzete Pitagorasz tétele szerint $\displaystyle (x+z+t)^2 - (x+z-t)^2 = (s+y)^2 - (s-y)^2$, amiből $\displaystyle \dfrac{x+z}{s}=\dfrac{y}{t}$. A párhuzamos szelők tételéből pedig $\displaystyle \dfrac{x}{t}=\dfrac{y+z}{s}$. E két utóbbi egyenlőséget összeszorozva: $\displaystyle \dfrac{x(x+z)}{st}=\dfrac{y(y+z)}{st}$, $\displaystyle x^2-y^2 + z(x-y)=0$, $\displaystyle (x-y)(x+y+z)=0$, azaz $\displaystyle x=y$, így $\displaystyle AT=x+z= y+z= VB$.

### Statistics:

 90 students sent a solution. 4 points: 57 students. 3 points: 21 students. 2 points: 2 students. 1 point: 3 students. 0 point: 7 students.

Problems in Mathematics of KöMaL, September 2017