Difference between revisions of "2011 AIME II Problems/Problem 13"
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==Problem== | ==Problem== | ||
Point <math>P</math> lies on the diagonal <math>AC</math> of [[square]] <math>ABCD</math> with <math>AP > CP</math>. Let <math>O_{1}</math> and <math>O_{2}</math> be the [[circumcenter]]s of triangles <math>ABP</math> and <math>CDP</math> respectively. Given that <math>AB = 12</math> and <math>\angle O_{1}PO_{2} = 120^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a + b</math>. | Point <math>P</math> lies on the diagonal <math>AC</math> of [[square]] <math>ABCD</math> with <math>AP > CP</math>. Let <math>O_{1}</math> and <math>O_{2}</math> be the [[circumcenter]]s of triangles <math>ABP</math> and <math>CDP</math> respectively. Given that <math>AB = 12</math> and <math>\angle O_{1}PO_{2} = 120^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a + b</math>. | ||
+ | |||
+ | ==Quickest Method of Solving== | ||
+ | This is a combination of Solutions 1 and 2. | ||
+ | |||
+ | First, draw <math>O_1P,O_2P,BP,DP</math>. Then, observe that <math>\angle BAP=45</math> implies that <math>\angle BO_1P=90</math>. So, <math>\triangle BO_1P</math> is a <math>45-90-45</math> triangle. Similarly, observe that <math>DO_2P</math> is too. So, a rotation of <math>\angle O_1PO_2</math> to <math>\angle BPO_2</math> adds <math>45</math> degrees. Then, moving to <math>BPD</math> subtracts <math>45</math> degrees. Hence, <math>\angle BPD=120</math>. Let the intersection of <math>BD</math> and <math>AC</math> be <math>Q</math>. Then <math>BQP</math> is a <math>30-90-60</math> triangle, hence <math>QP=\frac{6\sqrt{2}}{\sqrt{3}}</math> (We know that <math>BQ</math> is <math>6\sqrt{2}</math>), or <math>QP=2\sqrt{6}</math> Finally, <math>AP=QP+AQ=2\sqrt{6}+6\sqrt{2}=\sqrt{24}+\sqrt{72} \Rightarrow \boxed{096}</math> | ||
==Solution 1== | ==Solution 1== | ||
− | |||
Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>. | Denote the [[midpoint]] of <math>\overline{DC}</math> be <math>E</math> and the midpoint of <math>\overline{AB}</math> be <math>F</math>. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of <math>AB</math> and <math>CD</math> and these bisectors go through <math>E</math> and <math>F</math>. | ||
− | It is given that <math>\ | + | It is given that <math>\angle O_{1}PO_{2}=120^{\circ}</math>. Because <math>O_{1}P</math> and <math>O_{1}B</math> are [[radius|radii]] of the same circle, the have the same length. This is also true of <math>O_{2}P</math> and <math>O_{2}D</math>. Because <math>m\angle CAB=m\angle ACD=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</math> are isosceles right triangles. Using the given information above and symmetry, <math>m\angle DPB = 120^{\circ}</math>. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees. |
Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles <math>O_{1}BF</math> and <math>O_{2}DE</math> have measures of 30 degrees. Thus, both triangles <math>O_{1}BF</math> and <math>O_{2}DE</math> are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, <math>DO_{2}=BO_{1}=4\sqrt{3}</math>. Because of 45-45-90 right triangles, <math>PB=PD=4\sqrt{6}</math>. | ||
− | Now, using [[Law of Cosines]] on <math>\triangle ABP</math> and | + | Now, letting <math>x = AP</math> and using [[Law of Cosines]] on <math>\triangle ABP</math>, we have |
+ | |||
+ | <cmath>96=144+x^{2}-24x\frac{\sqrt{2}}{2}</cmath> | ||
+ | <cmath>0=x^{2}-12x\sqrt{2}+48</cmath> | ||
+ | |||
+ | Using the quadratic formula, we arrive at | ||
+ | |||
+ | <cmath>x = \sqrt{72} \pm \sqrt{24}</cmath> | ||
+ | |||
+ | Taking the positive root, <math>AP=\sqrt{72}+ \sqrt{24}</math> and the answer is thus <math>\framebox[1.5\width]{096.}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | This takes a slightly different route than Solution 1. | ||
+ | |||
+ | Solution 1 proves that <math>\angle{DPB}=120^{\circ}</math> and that <math>\overline{BP} = \overline{DP}</math>. | ||
+ | Construct diagonal <math>\overline{BD}</math> and using the two statements above it quickly becomes clear that <math>\angle{BDP} = \angle{DBP} = 30^{\circ}</math> by isosceles triangle base angles. | ||
+ | Let the midpoint of diagonal <math>\overline{AC}</math> be <math>M</math>, and since the diagonals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles. | ||
+ | Since <math>\overline{AB} = 12</math>, <math>\overline{AC} = \overline{BD} = 12\sqrt{2}</math> and <math>\overline{BM} = \overline{DM} = 6\sqrt{2}</math>. | ||
+ | 30-60-90 triangles' sides are in the ratio <math>1 : \sqrt{3} : 2</math>, so <math>\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}</math>. | ||
+ | <math>\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}</math>. | ||
+ | Hence, <math>72 + 24 = \framebox[1.5\width]{096}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Use vectors. In an <math>xy</math> plane, let <math>(-s,0)</math> be <math>A</math>, <math>(0,s)</math> be <math>B</math>, <math>(s,0)</math> be <math>C</math>, <math>(0,-s)</math> be <math>D</math>, and <math>(p,0)</math> be P, where <math>s=|AB|/\sqrt{2}=6\sqrt{2}</math>. It remains to find <math>p</math>. | ||
+ | |||
+ | The line <math>y=-x</math> is the [[perpendicular bisector]] of <math>AB</math> and <math>CD</math>, so <math>O_1</math> and <math>O_2</math> lies on the line. Now compute the [[perpendicular bisector]] of <math>AP</math>. The center has coordinate <math>(\frac{p-s}{2},0)</math>, and the segment is part of the <math>x</math>-axis, so the perpendicular bisector has equation <math>x=\frac{p-s}{2}</math>. Since <math>O_1</math> is the [[circumcenter]] of triangle <math>ABP</math>, it lies on the perpendicular bisector of both <math>AB</math> and <math>AP</math>, so | ||
+ | <cmath> | ||
+ | O_1=(\frac{p-s}{2},-\frac{p-s}{2}) | ||
+ | </cmath> | ||
+ | Similarly, | ||
+ | <cmath> | ||
+ | O_2=(\frac{p+s}{2},-\frac{p+s}{2}) | ||
+ | </cmath> | ||
+ | The relation <math>\angle O_1PO_2=120^\circ</math> can now be written using [[Vectors|dot product]] as | ||
+ | <cmath> | ||
+ | \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}| | ||
+ | </cmath> | ||
+ | Computation of both sides yields | ||
+ | <cmath> | ||
+ | \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2} | ||
+ | </cmath> | ||
+ | Solve for <math>p</math> gives <math>p=s/\sqrt{3}=2\sqrt{6}</math>, so <math>AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}</math>. The answer is 72+24<math>\Rightarrow\boxed{096}</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | Translate <math>\triangle{ABP}</math> so that the image of <math>AB</math> coincides <math>DC</math>. Let the image of <math>P</math> be <math>P’</math>. | ||
+ | |||
+ | <math>\angle{DPC}=\angle{CPB}</math> by symmetry, and <math>\angle{APB}=\angle{DP’C}</math> because translation preserves angles. Thus <math>\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ</math>. Therefore, quadrilateral <math>CPDP’</math> is cyclic. Thus the image of <math>O_1</math> coincides with <math>O_2</math>. | ||
+ | |||
+ | <math>O_1P</math> is parallel to <math>O_2P’</math> so <math>\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ</math>, so <math>\angle{PDP’}=60^\circ</math> and <math>\angle{PDC}=15^\circ</math>, thus <math>\angle{ADP}=75^{\circ}</math>. | ||
− | <math> | + | Let <math>M</math> be the foot of the perpendicular from <math>D</math> to <math>AC</math>. Then <math>\triangle{AMD}</math> is a 45-45-90 triangle and <math>\triangle{DMP}</math> is a 30-60-90 triangle. Thus |
− | <math> | + | <math>AM=6\sqrt{2}</math> and <math>MP=\frac{6\sqrt{2}}{\sqrt{3}}</math>. |
− | <math> | + | This gives us <math>AP=AM+MP=\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math> |
− | + | ==Solution 5== | |
− | <math> | + | Reflect <math>O_1</math> across <math>AP</math> to <math>O_1'</math>. By symmetry <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math> |
− | <math> | + | <math>\angle{DO_1’P}</math> = <math>2*\angle{DAP} = 90^\circ</math>, so <math>\angle{O_1’PD}=45^\circ</math> |
− | <math> | + | similarly <math>\angle{DO_2P}</math> = <math>2*\angle{DCP} = 90^\circ</math>, so <math>\angle{O_2PD}=45^\circ</math> |
− | <math> | + | Therefore <math>\angle{O_1’PO_2}=90^\circ</math>, so that <math>\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ</math> |
− | <math> | + | By symmetry, <math>\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ</math> |
+ | Therefore, since <math>O_1’</math> is the circumcenter of <math>\triangle{ADP}</math>, <math>\angle{ADP}</math> = <math>0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ</math> | ||
− | + | Therefore <math>\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ</math> | |
− | |||
− | ==Solution 2== | + | Using sine rule in <math>\triangle{ADP}</math>, <math>AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}</math>, and the answer is <math>72+24=\boxed{096}.</math> |
− | This | + | |
+ | By Kris17 | ||
+ | |||
+ | ==Solution 6 (Coordinates)== | ||
+ | Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set <math>A=(0, 12)</math>, <math>B=(12,12)</math>, <math>C=(12, 0)</math>, <math>D=(0, 0)</math>. Let this <math>P=(a, 12-a)</math> for some <math>a</math>. | ||
+ | |||
+ | We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that <math>O_{1}</math> is the intersection of <math>x=6</math> and, knowing the midpoint of <math>AP</math> to be <math>(\frac{a}{2}, \frac{12-a}{2})</math> and thus the equation to be <math>y=x+(12-a)</math>, we get <math>(6, 18-a)</math>. Likewise for <math>O_{2}</math> it's <math>(6, 6-a)</math>. Now what do we see? <math>O_{1}P=O_{2}P</math> (just look at the coordinates)! So both of those distances are <math>4\sqrt{3}</math>. Solving for <math>a</math> we get it to be <math>6+2\sqrt{3}</math>, since <math>AP>CP</math>. Multiply by <math>\sqrt{2}</math> because we are looking for <math>AP</math> to get the answer of <math>\boxed{096}</math>. | ||
+ | |||
+ | ==Solution 7 (Pure Angle Chasing)== | ||
+ | Let <math>\angle APD = \theta</math>. Then <math>\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}</math>. | ||
+ | Realize that because <math>O_1</math> is a circumcenter, <math>\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}</math>. Then <math>\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta</math>. | ||
+ | Now, because <math>P</math> lies on diagonal <math>AC</math>, <math>\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}</math>. | ||
+ | To finish, we look at <math>\triangle ADP</math>. Drop a perpendicular from <math>D</math> to <math>AP</math> at <math>E</math>. Then <math>\triangle ADE</math> is a <math>45-45-90</math> and <math>\triangle PDE</math> is a <math>30-60-90</math>. Therefore, <math>DE=EA=6\sqrt{2}, EP=2\sqrt{6}</math>, so <math>AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}</math>. <math>\blacksquare</math> ~msc | ||
+ | |||
+ | ==Solution 8 (similar to solution 4)== | ||
+ | Both <math>O_1</math> and <math>O_2</math> lie on the perpendicular bisector of <math>AB</math>. | ||
+ | |||
+ | '''Claim:''' <math>O_1O_2=12</math> and <math>O_1P=O_2P</math>. | ||
+ | |||
+ | ''Proof.'' Translate <math>O_1</math> and <math>P</math> <math>12</math> units down, and let their images be <math>O_1'</math> and <math>P'</math>, respectively. Note that <math>\triangle ABP\cong\triangle DCP'</math>. Additionally, <cmath>\angle CP'D = \angle BPA = 180^{\circ} - \angle BPC = 180^{\circ} - \angle CPD,</cmath> so <math>CPDP'</math> is cyclic. This means <math>O_1'</math> and <math>O_2</math> coincide, so <math>O_1O_2=12</math>. This also means the circumradii of both triangles are equal, so <math>O_1P=O_2P</math>. <math>\blacksquare</math>. | ||
+ | |||
+ | Let the perpendicular from <math>P</math> intersect <math>O_1O_2</math> at <math>X</math> and <math>AD</math> at <math>Y</math>. Since <math>\triangle O_1XP</math> is 30-60-90, <math>XP=\frac{6}{\sqrt{3}} = 2\sqrt3</math>. Since <math>YX=6</math>, <math>PY=6+2\sqrt3</math>, so <math>AP=6\sqrt2+2\sqrt6 = \sqrt{72}+\sqrt{24} \implies\boxed{96}</math>. | ||
+ | |||
+ | ~rayfish | ||
+ | |||
+ | ==Solution 9 Visual== | ||
+ | [[File:2011 AIME II 131.png|350px|right]] | ||
+ | <cmath>BP = DP, \angle PAB =\angle PCD = 45^\circ \implies O_{1}P =O_{2}P</cmath> by the Law of Sines. | ||
+ | |||
+ | <cmath>AB= CD, O_{1}A = O_{2}D = O_{1}B = O_{2}C \implies \triangle AO_{1}B = \triangle DO_{2}C.</cmath> They have translational symmetry. The translation vector is <cmath>\vec {AD} \implies O_{1}O_{2} = AD.</cmath> | ||
− | + | <cmath>\triangle PO_{1}O_{2} = \triangle O_{1}AB \implies \angle AO_{1}B = 120^\circ \implies</cmath> | |
− | + | <cmath>\angle APB = 60^\circ \implies \angle ABP = 180^\circ – 60^\circ – 45^\circ = 75^\circ.</cmath> | |
− | + | By the Law of Sines <cmath>AP = \frac {AB \cdot \sin 75^\circ}{\sin 60^\circ} = \frac {AB \cdot \sin (30^\circ + 45^\circ)} {\sin 60^\circ}</cmath> | |
− | + | <cmath>AP = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),</cmath> | |
− | 30 | + | <cmath>AP = \frac {AB}{\sqrt{2}} (1 + \frac {1}{\sqrt{3}}) = 6 \sqrt {2} + 2 \sqrt {6} \implies \boxed{\textbf{096}.}</cmath> |
− | + | '''vladimir.shelomovskii@gmail.com, vvsss''' | |
− | |||
==See also== | ==See also== | ||
Line 50: | Line 130: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 05:25, 9 September 2022
Contents
Problem
Point lies on the diagonal
of square
with
. Let
and
be the circumcenters of triangles
and
respectively. Given that
and
, then
, where
and
are positive integers. Find
.
Quickest Method of Solving
This is a combination of Solutions 1 and 2.
First, draw . Then, observe that
implies that
. So,
is a
triangle. Similarly, observe that
is too. So, a rotation of
to
adds
degrees. Then, moving to
subtracts
degrees. Hence,
. Let the intersection of
and
be
. Then
is a
triangle, hence
(We know that
is
), or
Finally,
Solution 1
Denote the midpoint of be
and the midpoint of
be
. Because they are the circumcenters, both Os lie on the perpendicular bisectors of
and
and these bisectors go through
and
.
It is given that . Because
and
are radii of the same circle, the have the same length. This is also true of
and
. Because
,
. Thus,
and
are isosceles right triangles. Using the given information above and symmetry,
. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.
Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles and
have measures of 30 degrees. Thus, both triangles
and
are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus,
. Because of 45-45-90 right triangles,
.
Now, letting and using Law of Cosines on
, we have
Using the quadratic formula, we arrive at
Taking the positive root, and the answer is thus
Solution 2
This takes a slightly different route than Solution 1.
Solution 1 proves that and that
.
Construct diagonal
and using the two statements above it quickly becomes clear that
by isosceles triangle base angles.
Let the midpoint of diagonal
be
, and since the diagonals are perpendicular, both triangle
and triangle
are 30-60-90 right triangles.
Since
,
and
.
30-60-90 triangles' sides are in the ratio
, so
.
.
Hence,
.
Solution 3
Use vectors. In an plane, let
be
,
be
,
be
,
be
, and
be P, where
. It remains to find
.
The line is the perpendicular bisector of
and
, so
and
lies on the line. Now compute the perpendicular bisector of
. The center has coordinate
, and the segment is part of the
-axis, so the perpendicular bisector has equation
. Since
is the circumcenter of triangle
, it lies on the perpendicular bisector of both
and
, so
Similarly,
The relation
can now be written using dot product as
Computation of both sides yields
Solve for
gives
, so
. The answer is 72+24
Solution 4
Translate so that the image of
coincides
. Let the image of
be
.
by symmetry, and
because translation preserves angles. Thus
. Therefore, quadrilateral
is cyclic. Thus the image of
coincides with
.
is parallel to
so
, so
and
, thus
.
Let be the foot of the perpendicular from
to
. Then
is a 45-45-90 triangle and
is a 30-60-90 triangle. Thus
and
.
This gives us , and the answer is
Solution 5
Reflect across
to
. By symmetry
is the circumcenter of
=
, so
similarly =
, so
Therefore , so that
By symmetry,
Therefore, since is the circumcenter of
,
=
Therefore
Using sine rule in ,
, and the answer is
By Kris17
Solution 6 (Coordinates)
Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set ,
,
,
. Let this
for some
.
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that is the intersection of
and, knowing the midpoint of
to be
and thus the equation to be
, we get
. Likewise for
it's
. Now what do we see?
(just look at the coordinates)! So both of those distances are
. Solving for
we get it to be
, since
. Multiply by
because we are looking for
to get the answer of
.
Solution 7 (Pure Angle Chasing)
Let . Then
.
Realize that because
is a circumcenter,
. Then
.
Now, because
lies on diagonal
,
.
To finish, we look at
. Drop a perpendicular from
to
at
. Then
is a
and
is a
. Therefore,
, so
.
~msc
Solution 8 (similar to solution 4)
Both and
lie on the perpendicular bisector of
.
Claim: and
.
Proof. Translate and
units down, and let their images be
and
, respectively. Note that
. Additionally,
so
is cyclic. This means
and
coincide, so
. This also means the circumradii of both triangles are equal, so
.
.
Let the perpendicular from intersect
at
and
at
. Since
is 30-60-90,
. Since
,
, so
.
~rayfish
Solution 9 Visual
by the Law of Sines.
They have translational symmetry. The translation vector is
By the Law of Sines
vladimir.shelomovskii@gmail.com, vvsss
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.