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USA Mathematical Talent Search Solutions To Problem 3/3/17

This problem asks the reader to find the value of CZ/AD + AX/BE + BY/CF for any acute triangle ABC on a circle, where points X, Y, Z are constructed such that AX is perpendicular to BC, BY is perpendicular to AC, and CZ is perpendicular to AB. The summary provides two solutions: 1) Using similarity of triangles and properties of orthocenters, the expression can be written in terms of ratios of triangle areas, yielding a value of 4. 2) Using the Power of a Point theorem and trigonometric identities involving tangent, the expression can be shown to always equal 4 for any acute triangle ABC.

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52 views2 pages

USA Mathematical Talent Search Solutions To Problem 3/3/17

This problem asks the reader to find the value of CZ/AD + AX/BE + BY/CF for any acute triangle ABC on a circle, where points X, Y, Z are constructed such that AX is perpendicular to BC, BY is perpendicular to AC, and CZ is perpendicular to AB. The summary provides two solutions: 1) Using similarity of triangles and properties of orthocenters, the expression can be written in terms of ratios of triangle areas, yielding a value of 4. 2) Using the Power of a Point theorem and trigonometric identities involving tangent, the expression can be shown to always equal 4 for any acute triangle ABC.

Uploaded by

Arsy
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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USA Mathematical Talent Search

Solutions to Problem 3/3/17


www.usamts.org

3/3/17. Points A, B, and C are on a circle such that 4ABC is


an acute triangle. X, Y , and Z are on the circle such that AX is
perpendicular to BC at D, BY is perpendicular to AC at E, and CZ
is perpendicular to AB at F . Find the value of
CZ
AX BY
+
+
,
AD BE CF

A
Y
E
F

Z
B

and prove that this value is the same for all possible A, B, C on the
circle such that 4ABC is acute.

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Credit This problem was proposed by Naoki Sato.


Comments This geometry problem can be solved by recognizing that the given ratios can
be expressed as ratios of certain areas, and using the fundamental result that HD = DX,
where H is the orthocenter of triangle ABC. A solution using power of a point is also
possible. Solutions edited by Naoki Sato.
Solution 1 by: Justin Hsu (11/CA)
Let H be the orthocenter of 4ABC. First, 4BHD is similar to 4BCE, since they are
both right triangles and they share CBE, so BCE = BHD. Also, BXA = BCA =
BHD, since they both are inscribed angles that intercept the same arc BA. Now, 4BXH
is isosceles, which means that BD is the perpendicular bisector of segment HX. Therefore,
4BDH 4BDX, and HD = DX. Similarily, this can be extended to the other sides of
the triangle to show that HE = EY and HF = F Z.
Now,
CZ
AD + DX BE + EY
CF + F Z
AX BY
+
+
=
+
+
AD BE CF
AD
BE
CF
DX
EY
FZ
=1+
+1+
+1+
AX
BY
CZ
HD HE HF
=3+
+
+
.
AD
BE
CF
But each fraction is a ratio between the altitudes of two triangles with the same base, so
we can rewrite this sum in terms of area, where [ABC] denotes the area of 4ABC:
3+

HD HE HF
[HBC] [HCA] [HAB]
+
+
=3+
+
+
AD
BE
CF
[ABC]
[ABC]
[ABC]
[ABC]
=3+
[ABC]
= 3 + 1 = 4.

USA Mathematical Talent Search


Solutions to Problem 3/3/17
www.usamts.org

Solution 2 by: James Sundstrom (11/NJ)


By the Power of a Point Theorem,
AD DX = BD CD.
Therefore,
BD CD
DX
=

= cot B cot C.
AD
AD AD
Similarly,
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EY
= cot C cot A,
BE
FZ
= cot A cot B.
CF

We can calculate
CZ
AD DX BE EY
CF
FZ
AX BY
+
+
=
+
+
+
+
+
AD BE CF
AD
AD
BE BE CF
CF
DX EY
FZ
=3+
+
+
AD
BE CF
= 3 + cot B cot C + cot C cot A + cot A cot B
tan A + tan B + tan C
.
=3+
tan A tan B tan C
We claim that
tan A + tan B + tan C = tan A tan B tan C
for all acute triangles 4ABC (acuteness of 4ABC means that tan A, tan B, and tan C
exist). [Ed: As the following argument shows, this identity holds for all triangles ABC
where both sides are defined.]
We have that
tan C = tan( A B) = tan(A B) = tan(A + B),
and
tan(A + B) =

tan A + tan B
,
1 tan A tan B

so

tan A + tan B
.
1 tan A tan B
This can be re-arranged to become tan A + tan B + tan C = tan A tan B tan C.
tan C =

Hence,
AX BY
CZ
+
+
= 3 + 1 = 4.
AD BE CF

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