Japan “Sangaku” Research Institute

Problem Let $D$ be the midpoint of the side $AB$ of $△ABC$, and the point $E$ be on the side $AC$ so that $AE∶EC=2∶1$. Moreover, let $O$ be the intersection of $CD$ and $BE$. Then, $$BE∶OE=4∶1.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem If the medians $BE$ and $CF$ of $△ABC$ are extended so that $BE=EG$ and $CF=FH$, then $G, \ A$ and $H$ are collinear. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution For $△AFH$ and $△BFC$, $$∠AFH=∠BFC, \qquad AF=BF,…

Problem The points $B$ and $C$ are on the same side of the line $XY$, and $A$ is on the opposite side. When the sum of the distances from $B$ and $C$ to $XY$ is equal to the distance from $A$ to $XY$,   $XY$ passes through the center of gravity of $△ABC$. $$ $$ $$ $$ $\downarrow$…

Problem If the feet of perpendicular lines drawn from vertices $A, \ B$, and $C$ to any line $XY$ passing through the center of gravity $G$ of $△ABC$ are $P, \ Q$, and $R$ respectively, then $$AP=BQ+CR.$$ However, $A$ is on the opposite side of $XY$ from $B$ and $C$. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$…

Problem The sum of the lengths of the perpendicular lines $AL, \ BM$, and $CN$ drawn from the vertices of $△ABC$ to a line outside the triangle is equal to three times the length of the perpendicular line $GP$ drawn from the center of gravity $G$ of the triangle to the line outside the triangle.…

Problem Take the points $E$ and $F$ on the sides $AB$ and $AC$ of $△ABC$ respectively, and $BE$ and $CF$ intersect at $G$. If $2GE=GB$ and $2GF=GC$,  then $G$ is the center of gravity of $△ABC$. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem In $△ABC$, let the two medians be $BE$ and $CF$.   If $AB>AC$, then $$BE>CF.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let the other median be $AD$, and the center of gravity of $△ABC$ be $G$. $△ABD$…

Problem The medians of the three sides of a triangle intersect at a point located $\frac{2}{3}$ of the way from each vertex. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let the three medians of $△ABC$ be $AM, \…