Inside the rectangle $ABCD$ is taken a point $M$ such that $\angle BMC + \angle AMD = 180^o$. Determine the sum of the angles $BCM$ and $DAM$.
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Tags: geometry, rectangle, angles
eduD_looC
27.07.2021 16:10
\bump \bump
Bowser498
27.07.2021 16:26
Is it not $90^\circ$?
Nahul2005
27.07.2021 17:47
is it $180^\circ$
vanstraelen
27.07.2021 19:04
Let $A(0,0),B(b,0),C(b,c),D(0,c)$; choose $M(m,n)$. $\tan \angle BMC=\tan \alpha=-\frac{c(m-b)}{(m-b)^{2}+n(n-c)}$ and $\tan \angle AMD=\tan \beta=\frac{cm}{m^{2}+n(n-c)}$. $\alpha+\beta=180^{\circ}$, then $\tan \beta=\tan(180^{\circ}-\alpha=-\tan \alpha$, giving $\frac{cm}{m^{2}+n(n-c)}=\frac{c(m-b)}{(m-b)^{2}+n(n-c)}$, after calculating $(m-b)m=n(n-c)$. The five points lie on an hyperbola $x^{2}-y^{2}-bx+cy=0$. $\tan \angle MAD=\tan y=\frac{m}{n}$ and $\tan \angle MCB=\tan x=\frac{m-b}{n-c}$. $\tan(x+y)=\frac{\frac{m-b}{n-c}+\frac{m}{n}}{1-\frac{m-b}{n-c} \cdot \frac{m}{n}} \Rightarrow x+y=90^{\circ}$.
Mathelete101
06.07.2022 23:01
bump... Is there a clean solution? I think reflecting M over the midsegment of the rectangle could be useful...
Bowser498
07.07.2022 07:38
At the time, I had a fakesolve, and still do.
Assume $ABCD$ is a square (which must also be a rectangle). Let $M$ be the center of this square. Then the diagonals create four right angles at the center; two of these angles: $\angle{BMC}$ and $\angle{AMD}$, must also be $90^\circ$, so their sum is $180^\circ$, which is true. But then this implies that $\angle{BCM}=\angle{DAM}=45^\circ$ by isosceles triangles, so their sum is $\boxed{90^\circ}$.
Mathelete101
07.07.2022 07:48
ok i finally found a solution which I think should work
Reflect M over the midsegment of ABCD to get M'. Let $\angle AMD = \theta$. Let the intersection point of M'A and BM be X and let the intersection points of M'D and MC be Y. Note that by the properties of the reflections $MX$ and $M'X$ have equal lengths. Therefore, $\angle MXM' = \theta$. It follows that $\angle XAB=90 - \frac{\theta}{2}$. This means that $\angle M'AD= \angle M'DA$. This means that $M'A=MD$ and by symmetry, $M'B=M'C$. We can then find the answer of $\boxed{90}$ from angle chasing.
fuzimiao2013
07.07.2022 08:06
[asy][asy]
draw((0,0)--(5,0)--(5,4)--(0,4)--(0,0)--(1,2)--(0,4)--(5,4)--(1,2)--(5,0));
draw((0,0)--(-4,2)--(0,4)--(1,2)--(-4,2));
label((0,0), "$D$", SW);
label((5,0), "$C$", SE);
label((5,4), "$B$", NW);
label((0,4), "$A$", N);
label((1,2), "$M$", S*2);
label((-4,2), "$M'$", W);[/asy][/asy]
Translate $M$ as such. Then obviously, $AMDM'$ is cyclic and $MM'\perp AD$ since $M\to M'$ is a horizontal translation so $\angle BCM + \angle DAM = \angle M'DA + \angle DAM = \angle M'MA + \angle DAM = 180^{\circ} - 90^{\circ} = 90^{\circ}$.