340. Geometry problems for middle school.
by Virgil Nicula, Apr 13, 2012, 10:50 PM
PP1. Let 
with the circumcircle 
, the incircle 
and the 
-excircle 
. Denote 
and 
.
Construct
and 
so that 
. Prove that 
and 
.
Proof. Prove easily that
.
Remark that
.
PP2. Let with 
and 
so that 
.
Let the circumcircle of 
and 
. Prove that 
.
Proof. Suppose w.l.o.g.
. Let midpoints 
of ![$[BF]\ ,\ [BG]$](//latex.artofproblemsolving.com/a/f/b/afb481d2432f86b7a24455a6fb6eca61974b7f3c.png)
and 
. Using the power of w.r.t. obtain that 
. Therefore, 
.
PP3 (clasa a VII - a). Let with 
. Denote the -bisector 
, where 
. Prove that 
.
Proof. Construct the equilateral
, 
so that 
separates 
, 
and 
separates 
, . Thus, 
and
.
In conclusion,
.
Remarks (clasa a IX - a).
Apply the generalized Pythagoras' theorem to the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
of 
.
Apply the Pompeiu's theorem (a particular case of the Ptolemy's theorem) to the cyclical quadrilateral 
.
Using the identity 
obtain that 
, i.e. .
PP4. Let An isosceles trapezoid with the incircle
which touches it at 
. Prove that ![$[LAN]=[LBT]=\frac 34\cdot [MNLT]$](//latex.artofproblemsolving.com/0/c/1/0c197bc8cf34a1c6e209781cd3075fc256305bbe.png)
.
Proof 1. Let
. Is well-known that 
and 
. Thus, 
(Brianchon's theorem) and
and ![$\odot\begin{array}{ccccccccc}
\nearrow & [LBT] & = & [LBP]+[LPT] & = & [LNP]+[LPT] & = & [LNPT] & \searrow\\\\
\searrow & [LAN] & = & [LAP]+[LPN] & = & [LTP]+[LPN] & = & [LNPT] & \nearrow\end{array}\odot$](//latex.artofproblemsolving.com/0/d/0/0d054b1a1aa55dc4471f0f4d63fe063fffb93f01.png)
![$[LBT]=[LAN]=[LNPT]=$](//latex.artofproblemsolving.com/a/f/d/afdc78037d24ebdc6d832aca9e90ef04a290e404.png)
![$\cdot \frac 12\cdot \frac 34\cdot ML\cdot NT=\frac 34\cdot [MNLT]$](//latex.artofproblemsolving.com/0/3/c/03c9d9c54f3ac526d90ddd47f0a60b4f90f9b403.png)
.
Proof 2. I"ll apply the well-known property: if
, where 
is between 
and 
, then for any 
and for any 
we have
![$[XAYB]=\frac 12\cdot AB\cdot \delta$](//latex.artofproblemsolving.com/7/4/7/747890a3f127ed84b87c5be08a4ddeade460ae1f.png)
(constant), where 
is the distance between 
, 
. Therefore, ![$[NAL]=[TBL]=\frac 12\cdot PL\cdot NT=\frac 12\cdot \frac 34\cdot ML\cdot NT=\frac 34\cdot [MNLT]$](//latex.artofproblemsolving.com/3/e/8/3e8dfa824a079a72bf1e2cc11d3b9351295fdcd7.png)
.
Remark. Consider an trapezoid
with 
and 
. Suppose w.l.o.g. 
and denote 
so that 
and 
. Let ![$[ABCD]=S$](//latex.artofproblemsolving.com/0/f/8/0f86c6b9da8a5fdcf8087f0d8f6a15023033c46d.png)
and 
. Prove easily that the following properties: ![$[AIB]=[CID]\le\frac S4\ ;\ [BIC]\cdot [AID]=$](//latex.artofproblemsolving.com/7/9/1/7913b67ba24fd5d7558dd93308540a5b64629da2.png)
![$[AIB]^2\ ;$](//latex.artofproblemsolving.com/f/6/4/f6453124632909e308bb8fb8be1e08300b8d9681.png)
and 
, i.e. 
is the harmonical mean between and .
PP5. Let line
, points 
in this order so that 
and point 
so that 
. Prove that 
.
Proof 1. Apply directy Steiner's theorem, where
, 
are isogonals in 
and will obtain 
. Let projection of 
on . Thus,
. Otherwise, can pass through the proof of the Steiner's theorem. In this case
construct parallelogram
and 
, 
. Thus, 
and 
. In conclusion, 
and 
a.s.o.
Proof 2.
cut again circumcircle of 
in . Thus, 
.
Proof 3. Let
and 
. Thus, 
.
Proof 4. I"use same notations. Thus,
.
Proof 5. Denote
and 
. Therefore, 
. In conclusion,
and 
.
PP6. Let semicircle
with diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and a rectangle so that 
is tangent to .
For
denote 
such that 
. Suppose that distancies of 
to is 
respectively and 
. Prove that 
. In particular, 
.
Proof. Denote the projections
of 
respectively on the line . Thus, 
and 
. Prove easily that 
and 
. In this case 
and prove easily that 
and 
.
An easy extension. Let the semicircle with the diameter . Construct the rectangle so that is tangent to .
For denote so that . Suppose that the distancies of to is respectively.
Denote
. Prove that 
.
Proof.
. Let midpoint of ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and projection 
of 
on 
. Thus, polygon 
is inscribed in circle with diameter ![$[OD]\implies$](//latex.artofproblemsolving.com/9/9/b/99b11abce8461cbe3099c4e2525b8616d759994c.png)
. Since
obtain 
![$MR^2=OM^2-OR^2=a^2\left[1-(\cos\phi -\sin\phi )^2\right]=a^2\sin2\phi\implies$](//latex.artofproblemsolving.com/f/9/c/f9c308d6f352082a4942d082185b3934cc3a72f8.png)
.Denote the projections of on , i.e.
and . Thus, 
![$\left[\sqrt{a^2-m^2}+\sqrt{a^2-n^2}\right]^2+(m-n)^2=$](//latex.artofproblemsolving.com/d/d/9/dd9d40877ee19ebed88e83dfbc00d22d5cc13fc2.png)
. Let projection of 
on the line 
. Thus,
Show easily that
and for 
get 
.
PP7. Let be a square with 
and let 
be a circle. For a mobile point ![$M\in \overarc[]{BD}$](//latex.artofproblemsolving.com/d/8/3/d8356daabb451b8bc04d46c06eb38035585f03d1.png)
denote 
.
Find the position of
so that 
, more exactly ascertain the common value of the equal angles.
Proof 1. Denote
. Observe that 
, 
and 
. Apply the teorem of Sines in
.
Proof 2. Since
obtain that 
is cylically 
is equilateral 
.
PP8. Let a rectangle with 
and circles 
so that 
is exterior
tangent to
and the circle is exterior tangent to 
. Denote 
which belongs to inside of and which is exterior tangent to 
. Prove that 
.
Proof. Let
and 
. Thus, 
and
In conclusion,
and 
.
Extension. Let a rectangle with 
and 
so that 
. Denote the midpoints 
, 
of , respectively and 
, 
so that
, i.e. 
. Denote 
so that is exterior tangent to and is exterior
tangent to . Let 
which belongs to inside of and which is exterior tangent to . Prove that 
and 
.
Proof. Let
. Thus, 
and 
. In conclusion, 
![$\left[x^2+(a+r)x-2ar\right]^2\iff$](//latex.artofproblemsolving.com/1/8/0/180cb1beec0728512bad70f13aaff4578aefd602.png)
![$\left[6ar-\left(a^2+r^2\right)\right]x^2+4ar(a+r)x-4a^2r^2=0\iff$](//latex.artofproblemsolving.com/e/e/c/eec5f331bd5c9fef10f2a6297cb3694e2a9cd8dc.png)
![$\frac {2ar\left[2\sqrt{2ar}-(a+r)\right]}{6ar-\left(a^2+r^2\right)}\implies$](//latex.artofproblemsolving.com/7/0/e/70eec0c64ee8862a022d0f1338e3736243df70ce.png)
. Observe that 
.
PP9. Let be a semicircle with the diameter . For a point 
with ![$\overarc[]{BT}<\overarc[]{TC}$](//latex.artofproblemsolving.com/1/d/9/1d948308a872ebccd5f23e7ecfe52c24e740e574.png)
denote the points 
and 
such that
. Denote 
. Prove that 
.
Proof.
. Apply an well-known relation (prove easily)
. But 
.
Therefore,
.
Remark.
is the bisector of 
. Thus, 
a.s.o.
PP10. Let and 
and 
so that 
. Prove that 
.
Proof. Let
, i.e. 
. Thus, 
.
In conclusion, 
.
PP11. Let-right triangle, the point 
and the midpoint of ![$[BD]$](//latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
so that 
. Prove that 
.
Proof 1. Denote w.l.o.g.
, 
and so that 
. Thus, 
. Therefore, 
and 
. In conclusion,
.
Proof 2. Apply trigonometric form of Ceva's theorem to and 
.
Proof 3.
. Thus, 
.
PP12. Let be a trapezoid, where 
and 
. For two points 
so that 
and 
.
Prove that
. With other words, ![$\boxed{MA\cdot [BCD]+MD\cdot [BCA]=AD\cdot [BCM]}$](//latex.artofproblemsolving.com/b/d/b/bdb78e548475050df83be323e100156c9c804f4b.png)
, where ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
.
Proof. Suppose w.l.o.g.
and let 
so that 
and 
.
Thus,
. Hence 
.
PP13. Let a trapezoid where 
and 
are the midpoints of ![$[AB]\ ,\ [CD]\ ,\ [AC]\ ,\ [BD]$](//latex.artofproblemsolving.com/e/5/4/e549b2d710731361593f6ce87439e1bd6ca4e215.png)
. Prove that 
.
Proof. Let be projections of 
on and 
where 
. The relations 
are well-known. So![$\left\{\begin{array}{cccc}
BD^2-AC^2=\left[h^2+(b+v)^2\right]-\left[h^2+(b+u)^2\right] & = & (v-u)(v+u+2b)=(v-u)(a+b)\\\\
CD^2-AB^2=\left(h^2+v^2\right)-\left(h^2+u^2\right)=v^2-u^2 & = & (v-u)(v+u)=(v-u)(a-b)\end{array}\right\|$](//latex.artofproblemsolving.com/3/5/9/35910ad4bebdfb96e93a416e5c62ee0879406965.png)
. In conclusion, 
.
with the circumcircle 
, the incircle 
and the 
-excircle 
. Denote 
and 
.Construct
and 
so that 
. Prove that 
and 
.Proof. Prove easily that
.Remark that
.PP2. Let
with 
and 
so that 
.Let the circumcircle
of 
and 
. Prove that 
.Proof. Suppose w.l.o.g.
. Let midpoints 
of ![$[BF]\ ,\ [BG]$](http://latex.artofproblemsolving.com/a/f/b/afb481d2432f86b7a24455a6fb6eca61974b7f3c.png)
and 
. Using the power of w.r.t. obtain that 
. Therefore, 
.PP3 (clasa a VII - a). Let
with 
. Denote the -bisector 
, where 
. Prove that 
.Proof. Construct the equilateral
, 
so that 
separates 
, 
and 
separates 
, . Thus, 
and
.In conclusion,
.Remarks (clasa a IX - a).
Apply the generalized Pythagoras' theorem to the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
of 
. Apply the Pompeiu's theorem (a particular case of the Ptolemy's theorem) to the cyclical quadrilateral 
. Using the identity 
obtain that 
, i.e. .PP4. Let An isosceles trapezoid with the incircle
which touches it at 
. Prove that ![$[LAN]=[LBT]=\frac 34\cdot [MNLT]$](http://latex.artofproblemsolving.com/0/c/1/0c197bc8cf34a1c6e209781cd3075fc256305bbe.png)
.Proof 1. Let
. Is well-known that 
and 
. Thus, 
(Brianchon's theorem) and
and ![$\odot\begin{array}{ccccccccc}
\nearrow & [LBT] & = & [LBP]+[LPT] & = & [LNP]+[LPT] & = & [LNPT] & \searrow\\\\
\searrow & [LAN] & = & [LAP]+[LPN] & = & [LTP]+[LPN] & = & [LNPT] & \nearrow\end{array}\odot$](http://latex.artofproblemsolving.com/0/d/0/0d054b1a1aa55dc4471f0f4d63fe063fffb93f01.png)
![$[LBT]=[LAN]=[LNPT]=$](http://latex.artofproblemsolving.com/a/f/d/afdc78037d24ebdc6d832aca9e90ef04a290e404.png)
![$\cdot \frac 12\cdot \frac 34\cdot ML\cdot NT=\frac 34\cdot [MNLT]$](http://latex.artofproblemsolving.com/0/3/c/03c9d9c54f3ac526d90ddd47f0a60b4f90f9b403.png)
.Proof 2. I"ll apply the well-known property: if
, where 
is between 
and 
, then for any 
and for any 
we have![$[XAYB]=\frac 12\cdot AB\cdot \delta$](http://latex.artofproblemsolving.com/7/4/7/747890a3f127ed84b87c5be08a4ddeade460ae1f.png)
(constant), where 
is the distance between 
, 
. Therefore, ![$[NAL]=[TBL]=\frac 12\cdot PL\cdot NT=\frac 12\cdot \frac 34\cdot ML\cdot NT=\frac 34\cdot [MNLT]$](http://latex.artofproblemsolving.com/3/e/8/3e8dfa824a079a72bf1e2cc11d3b9351295fdcd7.png)
.Remark. Consider an trapezoid
with 
and 
. Suppose w.l.o.g. 
and denote 
so that 
and 
. Let ![$[ABCD]=S$](http://latex.artofproblemsolving.com/0/f/8/0f86c6b9da8a5fdcf8087f0d8f6a15023033c46d.png)
and 
. Prove easily that the following properties: ![$[AIB]=[CID]\le\frac S4\ ;\ [BIC]\cdot [AID]=$](http://latex.artofproblemsolving.com/7/9/1/7913b67ba24fd5d7558dd93308540a5b64629da2.png)
![$[AIB]^2\ ;$](http://latex.artofproblemsolving.com/f/6/4/f6453124632909e308bb8fb8be1e08300b8d9681.png)
and 
, i.e. 
is the harmonical mean between and .PP5. Let line
, points 
in this order so that 
and point 
so that 
. Prove that 
.Proof 1. Apply directy Steiner's theorem, where
, 
are isogonals in 
and will obtain 
. Let projection of 
on . Thus,
. Otherwise, can pass through the proof of the Steiner's theorem. In this caseconstruct parallelogram
and 
, 
. Thus, 
and 
. In conclusion, 
and 
a.s.o.Proof 2.
cut again circumcircle of 
in . Thus, 
.Proof 3. Let
and 
. Thus, 
.Proof 4. I"use same notations. Thus,
.Proof 5. Denote
and 
. Therefore, 
. In conclusion,
and 
.PP6. Let semicircle
with diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and a rectangle so that 
is tangent to .For
denote 
such that 
. Suppose that distancies of 
to
is 
respectively and 
. Prove that 
. In particular, 
.Proof. Denote the projections
of 
respectively on the line . Thus, 
and 
. Prove easily that 
and 
. In this case 
and prove easily that 
and 
. An easy extension. Let the semicircle
with the diameter . Construct the rectangle so that is tangent to .For
denote so that . Suppose that the distancies of to is respectively.Denote
. Prove that 
.Proof.
. Let midpoint of ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and projection 
of 
on 
. Thus, polygon 
is inscribed in circle with diameter ![$[OD]\implies$](http://latex.artofproblemsolving.com/9/9/b/99b11abce8461cbe3099c4e2525b8616d759994c.png)
. Since
obtain 
![$MR^2=OM^2-OR^2=a^2\left[1-(\cos\phi -\sin\phi )^2\right]=a^2\sin2\phi\implies$](http://latex.artofproblemsolving.com/f/9/c/f9c308d6f352082a4942d082185b3934cc3a72f8.png)
.Denote the projections of on , i.e. and
. Thus, 
![$\left[\sqrt{a^2-m^2}+\sqrt{a^2-n^2}\right]^2+(m-n)^2=$](http://latex.artofproblemsolving.com/d/d/9/dd9d40877ee19ebed88e83dfbc00d22d5cc13fc2.png)
. Let projection of 
on the line 
. Thus,
Show easily that
and for 
get 
.PP7. Let
be a square with 
and let 
be a circle. For a mobile point ![$M\in \overarc[]{BD}$](http://latex.artofproblemsolving.com/d/8/3/d8356daabb451b8bc04d46c06eb38035585f03d1.png)
denote 
.Find the position of
so that 
, more exactly ascertain the common value of the equal angles.Proof 1. Denote
. Observe that 
, 
and 
. Apply the teorem of Sines in
.Proof 2. Since
obtain that 
is cylically 
is equilateral 
.PP8. Let a rectangle
with 
and circles 
so that 
is exteriortangent to
and the circle is exterior tangent to 
. Denote 
which belongs to inside of and which is exterior tangent to 
. Prove that 
.Proof. Let
and 
. Thus, 
and
In conclusion,
and 
.Extension. Let a rectangle
with 
and 
so that 
. Denote the midpoints 
, 
of , respectively and 
, 
so that
, i.e. 
. Denote 
so that is exterior tangent to and is exteriortangent to
. Let 
which belongs to inside of and which is exterior tangent to . Prove that 
and 
.Proof. Let
. Thus, 
and 
. In conclusion, 
![$\left[x^2+(a+r)x-2ar\right]^2\iff$](http://latex.artofproblemsolving.com/1/8/0/180cb1beec0728512bad70f13aaff4578aefd602.png)
![$\left[6ar-\left(a^2+r^2\right)\right]x^2+4ar(a+r)x-4a^2r^2=0\iff$](http://latex.artofproblemsolving.com/e/e/c/eec5f331bd5c9fef10f2a6297cb3694e2a9cd8dc.png)
![$\frac {2ar\left[2\sqrt{2ar}-(a+r)\right]}{6ar-\left(a^2+r^2\right)}\implies$](http://latex.artofproblemsolving.com/7/0/e/70eec0c64ee8862a022d0f1338e3736243df70ce.png)
. Observe that 
.PP9. Let
be a semicircle with the diameter . For a point 
with ![$\overarc[]{BT}<\overarc[]{TC}$](http://latex.artofproblemsolving.com/1/d/9/1d948308a872ebccd5f23e7ecfe52c24e740e574.png)
denote the points 
and 
such that
. Denote 
. Prove that 
.Proof.
. Apply an well-known relation (prove easily)
. But 
.Therefore,
.Remark.
is the bisector of 
. Thus, 
a.s.o.PP10. Let
and 
and 
so that 
. Prove that 
.Proof. Let
, i.e. 
. Thus, 
.In conclusion,
.PP11. Let
-right triangle, the point 
and the midpoint of ![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
so that 
. Prove that 
.Proof 1. Denote w.l.o.g.
, 
and so that 
. Thus, 
. Therefore, 
and 
. In conclusion,
.Proof 2. Apply trigonometric form of Ceva's theorem to
and 
.Proof 3.
. Thus, 
.PP12. Let
be a trapezoid, where 
and 
. For two points 
so that 
and 
.Prove that
. With other words, ![$\boxed{MA\cdot [BCD]+MD\cdot [BCA]=AD\cdot [BCM]}$](http://latex.artofproblemsolving.com/b/d/b/bdb78e548475050df83be323e100156c9c804f4b.png)
, where ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
.Proof. Suppose w.l.o.g.
and let 
so that 
and 
.Thus,
. Hence 
.PP13. Let a trapezoid
where 
and 
are the midpoints of ![$[AB]\ ,\ [CD]\ ,\ [AC]\ ,\ [BD]$](http://latex.artofproblemsolving.com/e/5/4/e549b2d710731361593f6ce87439e1bd6ca4e215.png)
. Prove that 
. Proof. Let
be projections of 
on and 
where 
. The relations 
are well-known. So
![$\left\{\begin{array}{cccc}
BD^2-AC^2=\left[h^2+(b+v)^2\right]-\left[h^2+(b+u)^2\right] & = & (v-u)(v+u+2b)=(v-u)(a+b)\\\\
CD^2-AB^2=\left(h^2+v^2\right)-\left(h^2+u^2\right)=v^2-u^2 & = & (v-u)(v+u)=(v-u)(a-b)\end{array}\right\|$](http://latex.artofproblemsolving.com/3/5/9/35910ad4bebdfb96e93a416e5c62ee0879406965.png)
. In conclusion, 
. This post has been edited 301 times. Last edited by Virgil Nicula, Nov 26, 2015, 3:23 PM
August 2018
January 2018