Math, asked by techybuster, 15 hours ago

Let
A
B
C
be a right angled triangle with

C
=
90
°
. Let
P
be any point lying on the same side of
A
B
as
C
such that

A
P
B
=
60
°
. Let
O
1
and
O
2
be the circumcenter of
A
P
C

and
B
P
C
. Let
k
be a real number such that for all choices of
P
,
[
A
B
C
]


k
[
O
1
C
O
2
]

Find the smallest value of

k
3
+
k
2
+
k
+
1

Answers

Answered by negiiabhinav
0

NiHatt merko krna hi Nahi h

Answered by ankan2778
0

Answer:

We begin with the following lemma:

Lemma: Let XYZ be a triangle with ∠XYZ=90+α. Construct an isosceles triangle XEZ, externally on the side XZ, with base angle.

Then E is the circumcentre of ΔXYZ.

Proof of the Lemma: Draw ED ⊥XZ.

Then DE is the perpendicular bisector of XZ. We also observe that ∠XED=ZED=90−α .

Observe that E is on the perpendicular bisector of XZ.

Construct the circumcircle of XY Z. Draw perpendicular bisector of XY and let it meet DE in F.

Then F is thecircumcentre of ΔXYZ. Join XF.

Then ∠XFD=90−α.

But we know that ∠XED=90−α. Hence E = F.

Let r

1

,r

2

and r be the inradii of the triangles ABD, CBD and ABCrespectively. Join PD and DQ. Observe that ∠PDQ=90

o

. Hence

PQ

2

=PD

2

+DQ

2

=2r

1

2

+2r

2

2

:

Let s

1

= (AB + BD + DA)=2. Observe that BD = ca=b and AD =

AB

2

−BD

2

=

c

2

−(

b

ca

)

2

=c

2

/b.

This gives s

1

= cs=b. But r

1

=s

1

c = (c=b)(s b) = cr=b. Similarly, r

2

= ar=b.

Hence PQ

2

=2r

2

(

b

2

c

2

+a

2

)=2r

2

Consider ΔPIQ. Observe thatPIQ=90+(B=2)=135.

Hence PQsubtends 90

o

on the circumference of the circumcircle of ΔPIQ. But we have seen that ∠PDQ=90

o

.

Now construct a circle with PQ asdiameter.

Let it cut AC again in K.It follows that ∠PKQ = 90 and thepoints P, D, K, Q are concyclic.

We also notice ∠KPQ=∠KDQ=45

o

and∠PQK=∠PDA=45

o

.

Thus PKQ is an isosceles right-angled triangle with KP = KQ.

∴KP

2

+KQ

2

=PQ

2

=2r

2

and hence KP = KQ = r.

Now ∠PIQ=90+45and∠PKQ=245

o

=90

o

with KP = KQ = r.

Hence K is the circumcentre of ΔPIQ.

(Incidentally,

This also shows that KI = r and hence K is the point of contact of the incircle of ΔABC with AC.)

Solution 2:

Here we use computation to prove that the point of contact K of the incircle with AC is the circumcentre of ΔPIQ.

We show that KP = KQ = r.

Let r

1

andr

2

be the in radii of triangles ABD and CBD respectively.

Draw PL⊥ACandQM⊥AC.

If s

1

isthe semiperimeter of ΔABD, thenAL=s

1

BD.

But s

1

=

2

AB+BD+DA

,BD=

b

ca

,AD=

b

c

2

Hence s

1

= cs=b.

This gives r

1

=s

1

−c=cr/b,AL=s

1

BD=c(sa)=b.

Hence KL=AKAL=(s−a)−

b

c(s−a)

=

b

(b−c)(s−a)

.

We observe that 2r

2

=

2

(c+ab)

2

=

2

c

2

+a

2

+b

2

2bc2ab+2ca

=(b

2

ba−bc+ac)=(b−c)(b−a).

This gives

(s - a)(b- c) = (s -b + b -a)(b- c) = r(b -c) + (b -a)(b- c)

= r(b-c) + 2r

2

= r(b-c + c + a-b) = ra.

Thus KL = ra=b.

Finally, KP

2

=KL

2

+LP

2

=

b

2

r

2

a

2

+

b

2

r

2

+c

2

=r

2

.

Thus KP = r.

Similarly, KQ = r.

This gives KP = KI = KQ = r and therefore K is the circumcentre of ΔKIQ.

(Incidentally, this also shows that KL=ca=b= r

2

and KM = r

1

.)

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