inradius of PMSN equals to MP-MS (Singapore Junior 2010)

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inradius of PMSN equals to MP-MS (Singapore Junior 2010)

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#1 h Jul 11, 2019, 12:15 AM • 2 Y

Y by ImSh95, Adventure10

Let the diagonals of the square $ABCD$ intersect at $S$ and let $P$ be the midpoint of $AB$ . Let $M$ be the intersection of $AC$ and $PD$ and $N$ the intersection of $BD$ and $PC$ . A circle is incribed in the quadrilateral $PMSN$ . Prove that the radius of the circle is $MP- MS$ .

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#2 h Jun 29, 2023, 2:34 PM • 1 Y

Y by ImSh95

Can anyone help me forward about this problem
I have done much of this but stucked at the last step

First extend the line ps to meet at f at dc then let the angle <mds=theta so <adm=45-theta from a bit of angle chasing you will get <apm=45+theta applying sine rule in triangle apm we get and I forgot to mention that sides of the square i let was x

PM/sin(45)=x/2(sin(90-theta)) or we get PM =sin(45)x/sin90costheta
Or PM =X/√2cos(theta)

Secondly we solve for MS also but first we need to find DS we can see easily it is equal to =√2x/2 then we can apply tan theta in traingle mds we get tan theta=MS/DS ON A BIT OF SIMPLIFICATION WE get MS =√2xtantheta/2
And at last we subtract those 2 values we get
x/√2×tantheta
Can anyone solve further for the value of r i cannot get it

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#3 h Jun 29, 2023, 4:38 PM • 1 Y

Y by ImSh95

Given the square $ABCD\ :\ A(0,0),B(2,0),C(2,2),D(0,2)$ .
Then $S(1,1)$ and $P(1,0)$ .

The line $AC\ :\ y=x$ intersects the line $DP\ :\ y=-2(x-1)$ in the point $M(\frac{2}{3},\frac{2}{3})$ .
The line $BD\ :\ y=2-x$ intersects the line $CP\ :\ y=2(x-1)$ in the point $N(\frac{4}{3},\frac{2}{3})$ .
Distance $MP=\frac{\sqrt{\sqrt{5}}}{3}$ and distance $MS=\frac{\sqrt{2}}{3}$ .

Let be $O(1,t)$ the midpoint of the incircle.
Distance of $O$ to the line $AC\ :\ \frac{1}{\sqrt{2}}(x-y)=0$ equals $\frac{1}{\sqrt{2}}(1-t)=r$ .
Distance of $O$ to the line $DP\ :\ \frac{1}{\sqrt{5}}(2x+y-2)=0$ equals $\frac{1}{\sqrt{5}} \cdot t = r$ .
Solving $t=\sqrt{5}r$ and $r=\frac{\sqrt{5}-\sqrt{2}}{3}$ .

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#4 h Jun 30, 2023, 4:01 AM • 1 Y

Y by ImSh95

We have
Let AB= BC = CD = DA = a
Radius of incribed circle in the quadrilateral is :r
Area of PMSN = 2 area of PMS= r ( PM +MS )

theorem Thales
(MP/MD)=(AP/CD)=(MA/MC)=(1/2)=> (PM/PD )=(1/3)
S[PMS ]=(1/2)S[MDS
]=>S[MPNS ]=S[MDS]

theorem Menelaus wehave
(DS/DB).(PB/PA).(MA/MS)=1
=> (MA/MS )=2
=>(S[MDS]/S[DSA])=(1/3)
=> S[MDS]=(1/3)S[DSA]=(1/4).(1/3)S[ABCD]=(a^2 /12)
=> S[PMSN ]= (a^2/12)
=> r = (a^2 /12( MP +MS ))
so we proof that a^2 = 12 (MP^2 - MS^2 )
MP^2 =(PD^2 /9)=(a^2 +(a^2/4)/9 )=(5 /36)a^2
MS ^2 = (SA^2 /9)=((a^2/2)/9)=(a^2/18)
QED

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#5 h May 24, 2024, 3:11 PM

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an areas solution

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