2
$\begingroup$

A square sheet of paper $ABCD$ is so folded that $B$ falls on the midpoint $M$ of $CD$. In which ratio will the crease divide $BC$?

2 Answers 2

-1

Let the side of the square be 2.

enter image description here

$\triangle BXY \sim \triangle BMC$. (Corrected. Originally was quoted as $\triangle BXY \sim \triangle AMC$.)

Setting up the corresponding ratios, we get $BX = \dfrac 54$ and hence $CX = 2 - BX = \dfrac 34$.

Extras (Completely re-written with the old picture replaced)

enter image description here

If, after folding, B falls on M, then the crease must be the perpendicular bisector of BM.

Let that perpendicular bisecor cut DC extended, BC, BM and AD at P, Q, R, S respectively.

By Pythagoras theorem, $MR = \dfrac {\sqrt 5}{2}$.

From $\triangle BQR \sim \triangle BMC$, we get $BQ = \dfrac {5}{4}$ and $QC = \dfrac {3}{4}$. In addition, QR is also known.

From $\triangle PCQ \sim \triangle PRM$, we get $PC = 1.5$.

From $\triangle PCQ \sim \triangle PDS$, we get $DS = \dfrac {7}{4}$.

Hence, $AS = 2 - DS = \dfrac 14$ and $DS : SA = 7 : 1$.

  • 1
    Looks like you folded $A$ to $M$, not $B$ to $M$ as in the question. – 2017-01-14
  • 0
    @QuangHoang Thanks for pointing that out. Edited and the original has been left as extras. – 2017-01-15
  • 0
    This answer is wrong. – 2018-08-08
  • 0
    @MalayTheDynamo Besides the typo fixed, what else is considered as seriously wrong? Care to say a bit more? – 2018-08-09
  • 0
    @Mick The answer. The actual value is $\frac 53$. Which appears nowhere in your answer. – 2018-08-09
  • 0
    @MalayTheDynamo If $BX ( = \dfrac 54)$ and $CX ( = \dfrac 34)$ were found. The required answer follows immediately. Therefore, my answer should not be consider as "wrong". FYI, this site is not for for homework doing and we will only provide adequate hints and hope the OP can complete the rest of their work from that point on. – 2018-08-10
0

Let the crease be $QR$, with $Q$ on $BC$ and $R$ on $BM$. Then notice that $QR$ is the perpendicular bisector of $BM$. Then $\Delta BQR\cong\Delta MQR$.

WLOG, $QC=1,BQ=x$. Then $MC=\frac{1+x}2,MQ=x$.

Then by Pythagoras on $\Delta MCQ$, we have $\left(\frac{1+x}2\right)^2+1=x^2$. Solving gives $x=\frac53$.

Thus, the crease divides $BC$ in a ratio of $5:3$.

$\boxed{\tiny Z}$