Difference between revisions of "1974 USAMO Problems/Problem 5"
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Line 10: | Line 10: | ||
draw(A--C); | draw(A--C); | ||
− | label(" | + | label("$A$",A,N); |
label("$B$",B,ESE); | label("$B$",B,ESE); | ||
label("$C$",C,SW); | label("$C$",C,SW); | ||
+ | label("$D$",D,NE); | ||
label("$a$",(B+C)/2,S); | label("$a$",(B+C)/2,S); | ||
label("$b$",(C+A)/2,WNW); | label("$b$",(C+A)/2,WNW); | ||
Line 74: | Line 75: | ||
&= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . | &= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | But the area of triangle <math>PQR</math> is <math> | + | But the area of triangle <math>PQR</math> is <math>x^2 \sqrt{3}/4</math>. It follows that <math>u+v+w=x</math>, as desired. <math>\blacksquare</math> |
=== Solution 2 === | === Solution 2 === | ||
Line 144: | Line 145: | ||
</asy> | </asy> | ||
These three triangles fit together because <math>(A+60)+(B+60)+(C+60) = 360</math>. The result is an equilateral triangle of side length <math>u+v+w</math>. | These three triangles fit together because <math>(A+60)+(B+60)+(C+60) = 360</math>. The result is an equilateral triangle of side length <math>u+v+w</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | <asy> | ||
+ | size(300); | ||
+ | defaultpen(1); | ||
+ | pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); | ||
+ | pair RR=R+Q-P, MM= rotate(-60,Q)*M; | ||
+ | |||
+ | draw(P--R--RR--Q--P--M--MM--RR); | ||
+ | draw(Q--R--M--Q--MM--R); | ||
+ | |||
+ | label("$P$",P,W); | ||
+ | label("$Q$",Q,E); | ||
+ | label("$R$",R,W); | ||
+ | label("$M$",M,NW); | ||
+ | label("$R'$",RR,NE); | ||
+ | label("$M'$",MM,ESE); | ||
+ | </asy> | ||
+ | |||
+ | As in the first solution, we rotate and establish that <math>\triangle MRM' \cong \triangle ABC</math>. | ||
+ | |||
+ | Let <math>X</math> and <math>Y</math> be points on <math>\overline{RQ}</math> and <math>\overline{PQ}</math>, respectively, such that <math>M</math> lies on <math>\overline{XY}</math> and <math>\overline{XY}\parallel \overline{PR}</math>. We note that <math>m\angle RXM = 120^\circ </math>. The rotation then takes <math>Y</math> to <math>X</math>, so <math>m\angle RXM'=m\angle PYM = 120^\circ</math>. | ||
+ | It follows that <math>RX=BD=v</math>, <math>MX=AD=u</math>, <math>M'X=CD=w</math>. | ||
+ | |||
+ | Since <math>m\angle MXM' = 120^\circ </math> and <math>m\angle MQM' = 60^\circ </math>, <math>MXM'Q</math> is cyclic. | ||
+ | By Ptolemy's theorem, | ||
+ | <cmath>\begin{align*} | ||
+ | (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \\ | ||
+ | XQ &= MX + M'X \\ | ||
+ | &= u+w. | ||
+ | \end{align*}</cmath> | ||
+ | Finally, <math>RQ = RX+XQ = u+v+w</math>, as desired. | ||
+ | |||
+ | |||
+ | |||
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == See Also == |
− | {{USAMO box|year=1974|num-b=4|aftertext=|after=Last | + | {{USAMO box|year=1974|num-b=4|aftertext=|after=Last Question}} |
*<url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url> | *<url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url> | ||
− | + | {{MAA Notice}} | |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 18:57, 3 July 2013
Problem
Consider the two triangles and
shown in Figure 1. In
,
. Prove that
.
Solutions
Solution 1
We rotate figure by a clockwise angle of
about
to obtain figure
:
Evidently, is an equilateral triangle, so triangles
and
are congruent. Also, triangles
and
are congruent, since they are images of each other under rotations. Then
Then by symmetry,
But is composed of three smaller triangles. The one with sides
has area
. Therefore, the area of
is
Also, by the Law of Cosines on that small triangle of
,
, so by symmetry,
Therefore
But the area of triangle
is
. It follows that
, as desired.
Solution 2
Rotate
degrees clockwise about
to get
. Observe that
is equilateral, which means
. Also,
are collinear because
and
. The resulting
has side lengths
and the angle opposite side
has magnitude
.
If we perform the rotation about points
and
, we get two triangles. One has side lengths
and the angle opposite side
has magnitude
, and the other has side lengths
and the angle opposite side
has magnitude
.
These three triangles fit together because
. The result is an equilateral triangle of side length
.
Solution 3
As in the first solution, we rotate and establish that .
Let and
be points on
and
, respectively, such that
lies on
and
. We note that
. The rotation then takes
to
, so
.
It follows that
,
,
.
Since and
,
is cyclic.
By Ptolemy's theorem,
Finally,
, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url>
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.