Japan SANGAKU Research Institute

Problem If a line is drawn from vertex $A$ of square $ABCD$ through a point $X$ on BC and a point $Y$ on the extension of $DC$, then $$AX+AY>2AC.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let $M$…

Problem If squares are constructed externally on the sides $AB, \ BC, \ CD$, and $DA$ of a convex quadrilateral $ABCD$, and their centers are denoted by $L, \ M, \ N$, and $P$, respectively, then the segments $LN$ and $PM$ are equal in length and perpendicular. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$…

Problem Draw squares $ABDE$ and $ACFG$ externally on sides $AB$ and $AC$ of triangle $ABC$, respectively, and let $P$ and $Q$ denote their centers. If $L$ and $M$ are the midpoints of $EG$ and $BC$, respectively, then the quadrilateral $LPMQ$ is a square. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem If squares $ABDE$ and $ACFG$ are constructed externally on sides $AB$ and $AC$ of triangle $ABC$, respectively, then $$BG = EC \qquad and \qquad BG ⟂ EC.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution   For triangles…

Problem If $M$ is the midpoint of side $AB$ of right triangle $ABC \ (∠C=∠R)$, and the squares $BCDE$ and $ACFG$ are constructed externally on sides $BC$ and $CA$, respectively, then $$CM=\frac{1}{2} DF \qquad and \qquad CM⊥DF.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…

Problem Construct squares $ABDE$ and $ACFG$ externally on triangle $ABC$, with sides $AB$ and $AC$ as their respective bases. Let $H$ be the foot of the perpendicular from $A$ to $BC$. Let $M$ be the point where line $HA$ intersects segment $EG$. Then, show that $$EM = MG.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$…