## Homework 6: Geometry Problem 1 If \( ABCD \) is a parallelogram, prove that \[ AB^2+BC^2+CD^2+DA^2=AC^2+BD^2.\] Let \( \overrightarrow a=\overrightarrow{AB}=\overrightarrow{DC} \) and \( \overrightarrow{b}=\overrightarrow{BC}=\overrightarrow{AD} \). Then \( \overrightarrow{AC}=\overrightarrow a+\overrightarrow b \) and \( \overrightarrow BD=\overrightarrow b-\overrightarrow a \). Now we have: \begin{eqnarray*}AC^2+BD^2&=&\left|\overrightarrow a+\overrightarrow b\right|^2+\left|-\overrightarrow a+\overrightarrow b\right|^2=\left|\overrightarrow a\right|^2+ \left|\overrightarrow b\right|^2+2\overrightarrow a\cdot\overrightarrow b+ \left|\overrightarrow a\right|^2+ \left|\overrightarrow b\right|^2-2\overrightarrow a\cdot\overrightarrow b\\ &=&AB^2+BC^2+CD^2+DA^2. \end{eqnarray*} The notation \( \overrightarrow a\cdot \overrightarrow b \) denotes the dot product of the vectors \( \overrightarrow a \) and \( \overrightarrow b \), i.e. \( \overrightarrow a\cdot \overrightarrow b=\left|\overrightarrow a\right|\cdot \left|\overrightarrow b\right|\cdot\cos\alpha \), where \( \alpha \) is the angle between \( \overrightarrow a \) and \( \overrightarrow b \). Problem 2 Let \( ABC \) be a triangle such that \( \angle CAB=\angle CBA=\alpha \). Assume that the incircle of the triangle touches the sides \( AB \) and \( AC \) in \( C^{\prime} \) and \( B^{\prime} \) respectively. Assume that \( I \) is the incenter of \( \triangle ABC \) and that \( AC’=x \). Find an expression, in terms of \( x \) and \( \alpha \), for the area of the curvilinear triangle bounded by the lines \( AB \) and \( AC \) and the arc of the incircle between \( C’ \) and \( B’ \). We have that \( IC’=IB’= x\tan\frac{\alpha}2 \), hence the area of the quadrilateral \( AC^{\prime}IB^{\prime} \) is equal to \( S_{AC^{\prime}IB^{\prime}}=2S_{\triangle AC^{\prime}I}=a^2\tan\frac{\alpha}2 \). The area of the circular region between \( IB^{\prime} \) and \( IC^{\prime} \) is equal to \( \frac12 IC^{\prime 2}\cdot \angle B^{\prime}IC^{\prime}=\frac12a^2\tan^2\frac{\alpha}2\cdot \left(\pi-\alpha\right) \). Therefore the required area satisfies \[ S=a^2\tan\frac{\alpha}2-\frac12a^2\tan^2\frac{\alpha}2\cdot \left(\pi-\alpha\right).\] Problem 3 Assume that a square is inscribed in a triangle such that all three vertices of the square belong to the edges of the triangle. Show that the area of the square is at most half of the area of the triangle. Let \( ABC \) be the triangle and let \( PQRS \) be the square such that \( P, Q\in BC \), \( R\in AC \), and \( S\in AB \). Let us denote by \( a \) the length of \( BC \), and by \( h \) the length of the altitude from \( A \) to \( BC \). Let \( x=PQ \) be the length of the edge of the square. Then \( S_{\triangle SRA}=\frac12 (h-x)\cdot x \), \( S_{\triangle SBP}=\frac12 x\cdot |BP| \), and \( S_{\triangle QCR}=\frac12 x\cdot |QC| \). Therefore \[ S_{PQRS}=\frac{hc-(h-x)x-x\left(|BP|+|QC|\right)}2= \frac{hc-(h-x)x-x(c-x)}2.\] We need to prove that \( \frac{hc-(h-x)x-x(c-x)}2\geq x^2 \). This is equivalent to \( -x^2-(h+c)x+hc\leq 0 \), or, \( -(c-x)(h-x)\leq 0 \) which is true. Problem 4 Consider an infinite sequence of circles \( (C_k)_{k=0}^{\infty} \) defined in the following way: The circle \( C_0 \) has center \( (1,1) \) and is tangent to the \( x \) and \( y \) axis. For \( k\geq 0 \) the circle \( C_{k+1} \) is defined to be the circle externally tangent to \( C_k \), tangent to the \( x \) axis, and with the center on the segment connecting \( (1,1) \) to \( (2,0) \). Denote by \( d_k \) the radius of \( C_k \). Evaluate \( \displaystyle \sum_{k=0}^{\infty} d_k \). Let \( A_k \) be the center of \( C_k \), \( B_k \) the point of tangency of \( C_k \) with the \( x \)-axis, \( D_k \) the point of tangency of \( C_k \) and \( C_{k+1} \), and \( P_k \) the feet of perpendicular from \( B_k \) to the line segment \( A_kB_k \). Then \( A_kA_{k+1}=d_k+d_{k+1} \), while \( A_kP_k=d_k-d_{k+1} \). Since \( \angle P_kA_kA_{k+1}=45^{\circ} \) we have that \( A_kA_{k+1}=A_kP_k\cdot \sqrt 2 \), hence \[ d_k+d_{k+1}=\sqrt 2\cdot \left(d_k-d_{k+1}\right)\quad\quad\quad \Rightarrow \quad\quad\quad d_{k+1}=\frac{\sqrt 2-1}{\sqrt2 +1}d_k=\left(\sqrt 2-1\right)^2d_k.\] Thus \( \displaystyle \sum_{k=0}^{\infty} d_k=\sum_{k=0}^{\infty}\left(\sqrt 2-1\right)^{2k}=\frac1{1-\left(\sqrt 2-1\right)^{2}}=\frac1{2\sqrt2 -2}=\frac{\sqrt 2+1}2 \). Problem 5 Every point in the plane is painted in blue or red. Show that there are either two blue points at the distance exactly one, or four colinear red points \( R_1 \), \( R_2 \), \( R_3 \), and \( R_4 \) such that \( R_1R_2=R_2R_3=R_3R_4=1 \). Suppose that there are no four colinear red points with the stated property. Let \( P \) be any blue point (there must be at least one blue point). Consider the circle \( k \) with center \( P \) and radius \( 1 \). If some point on the circle is blue, then the claim will follow immediately. Thus suppose that all points on the circle are red. Let \( ABCDEF \) be an arbitrary regular hexagon whose vertices belong to \( k \). Let \( X \) be intersection of \( AB \) with \( CD \), \( Y \) the intersection of \( CD \) with \( EF \), and \( Z \) the intersection point of \( EF \) with \( AB \). If at least two of the points \( X, Y, Z \) are red (say \( X \) and \( Y \)) then we can find four colinear red points (in our case, \( X, C, D, Y \)) with the stated property, and that would be a contradiction. Hence at least two of the points \( X, Y, Z \) are blue. Now, consider the rotation with the center \( P \) by which the points \( A \), \( B \), \( C \), \( D \), \( E \), \( F \), \( X \), \( Y \), \( Z \) are mapped to the points \( A^{\prime} \), \( B^{\prime} \), \( C^{\prime} \), \( D^{\prime} \), \( E^{\prime} \), \( F^{\prime} \), \( X^{\prime} \), \( Y^{\prime} \), \( Z^{\prime} \) such that \( XX^{\prime}=YY^{\prime}=ZZ^{\prime}=1 \). By the same argument we get that two of the points \( X^{\prime}, Y^{\prime}, Z^{\prime} \) are blue. The problem is now solved, since at least one of the segments \( XX^{\prime} \), \( YY^{\prime} \), \( ZZ^{\prime} \) has both ends painted in blue. Problem 6 Let \( ABCD \) be a rectangle. Let \( E \) be the foot of perpendicular from \( A \) to \( BD \). Let \( F \) be an arbitrary point of the diagonal \( BD \) between \( D \) and \( E \). Let \( G \) be the intersection of the line \( CF \) with the perpendicular from \( B \) to \( AF \). Let \( H \) be the intersection of the line \( BC \) with the perpendicular from \( G \) to \( BD \). Prove that \( \sphericalangle EGB=\sphericalangle EHB \). Let \( X \) be the intersection of lines \( BG \) and \( AE \). Since \( BX \perp AF \) and \( AX \perp BF \), \( X \) is the orthocenter of \( \triangle ABF \) hence \( FX \perp AB \Rightarrow FX\parallel BC \parallel AD \). It follows:\\ \[ \frac{EF}{ED}=\frac{FX}{DA}=\frac{FX}{BC}=\frac{GF}{GC}\;,\] (the first equality follows from \( \triangle EFX\sim \triangle EDA \), and the last from \( \triangle GFX\sim \triangle GCB \)). We conclude that \( GE\parallel CD \perp BH \). Since \( BE\perp GH \) it follows that \( E \) is the orthocenter of \( \triangle GHB \) and \( \sphericalangle EGB=\sphericalangle EHB \) follows easily (if we denote by \( H^{\prime} \) and \( G^{\prime} \) the intersections of \( GE \) and \( HE \), the triangles \( GG^{\prime}B \) and \( HH^{\prime}B \) are similar). Problem 7 Given a triangle \( ABC \), consider a circle \( k \) that is tangent to the lines \( AB \) and \( AC \) at \( B \) and \( P \). Let \( H \) be the foot of perpendicular from the center \( O \) of \( k \) to the line \( BC \), and let \( T \) be the intersection of \( OH \) and \( BP \). Prove that \( AT \) bisects the segment \( BC \). Let \( X \) be the intersection of \( AT \) and \( BC \), \( S \) the intersection of \( BC \) and \( AO \), and \( Y \) the intersection of \( BP \) and \( AO \). Since \( BY\perp SO \) and \( OH\perp BS \), \( T \) is the orthocenter of \( \triangle BOS \) and \( ST\perp BO \), hence \( ST\| AB \). Thus \[ ST:AB=TX:XA.\quad\quad\quad\quad\quad (1) \] However since \( AB=AP \) and the triangles \( SYT \) and \( AYP \) are similar we get \[ ST:AP=TY:YP. \quad\quad\quad\quad\quad (2)\] From (1) and (2) we now get \( XY\|AP \) and since \( Y \) is the midpoint of \( BP \), \( XY \) is the middle line of \( \triangle CBP \) and \( BX=XC \). Problem 8 Let \( K \) be the point \( K \) in the interior of the unit circle. The circle is divided into \( 8 \) regions using four lines passing through \( K \) such that each two adjacent lines form an angle of \( 45^{\circ} \), as shown in the picture. Four of these regions are colored such that no two colored regions share more than one point. Prove that the area of the colored regions is exactly half the area of the circle. Let us inscribe our circle in a square whose sides are parallel to 2 of these lines. If we color the regions of the square in the same way as the regions of the circle are colored we can easily prove that the total area of colored regions is equal to the area of non-colored regions for the square. Hence it remains to prove the equality of the areas of colored and uncolored regions that are outside the cirle. Now we will look at the pictures below. For each picture it is obvious that it has the same area of colored parts as the previous picture (as some colored parts are just moved from one location to the other). |

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